NCERT Solutions Class 9 Science Chapter 4 Describing Motion Around Us

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

CLASS 9 · SCIENCE

NCERT Solutions

Chapter 4: Describing Motion
Around Us

NCERT Textbook — Exploration

BOOK PAGES SECTIONS QUESTIONS MEDIUM

48 – 71 15 55 English

Solutions, notes, sample papers & more at 75 pages

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

CLASS 9 · SCIENCE · EXPLORATION

NCERT Solutions — Chapter 4: Describing Motion
Around Us
This chapter builds the language of kinematics: position with respect to a reference point, distance travelled
and displacement, average speed and average velocity, and average acceleration. It then shows how to
read the same motion off position–time and velocity–time graphs, derives the three kinematic equations for
constant acceleration, and closes with uniform circular motion.

TEXTBOOK BOOK PAGES

Exploration (Class 9) 48 – 71

SECTIONS QUESTIONS

15 55

MEDIUM

English

Think It Over — Page 48
Chapter opener

THINK IT OVER

Q1 How much distance should we maintain from the truck ahead to avoid a collision if
it suddenly applies the brakes?

Enough distance to cover the distance our own vehicle needs to stop — and that has two
separate parts.

stopping distance = reaction distance + braking distance

reaction distance = u × tr (we keep moving while the driver reacts)

braking distance = u² / (2|a|) (from v² = u² + 2as with v = 0)

Take a truck ahead and our car at u = 54 km h⁻¹ = 15 m s⁻¹, a driver reaction time tr ≈ 1 s and
braking that gives |a| = 4 m s⁻² on a dry road:

Page 1 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

reaction distance = 15 m s⁻¹ × 1 s = 15 m

braking distance = (15 m s⁻¹)² / (2 × 4 m s⁻²) = 225 / 8 m = 28.1 m

total = 15 m + 28.1 m = about 43 m

Why it happens: the truck ahead is also braking, so what really matters is the extra
distance we need compared with it. Our reaction time is pure loss — for that whole
second we travel at full speed with no braking at all. That is why the safe gap is
usually quoted as a time gap (keep 2–3 seconds behind the vehicle ahead) rather
than a fixed number of metres: a time gap automatically scales with speed.

Q2 Does this distance depend upon the speed with which we are moving?

Yes — and much more strongly than most people expect, because the braking part grows as the
square of the speed.

braking distance s = u² / (2|a|)

u = 15 m s⁻¹ (54 km h⁻¹) → s = 225 / 8 m = 28.1 m

u = 30 m s⁻¹ (108 km h⁻¹) → s = 900 / 8 m = 112.5 m

Doubling the speed does not double the braking distance — it makes it four times as long. The
reaction distance (u × tr) only doubles, so at high speed the braking term dominates completely.

Tip: the same formula explains why speed limits fall so sharply on wet roads, near
schools and in fog. A wet road lowers |a|, and s is inversely proportional to |a| —
halve the braking capacity and the stopping distance doubles at every speed.

In-text Questions — Page 51

Page 2 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Section 4.1.2 Distance travelled and displacement

Q1 Can these quantities ever be equal?

Yes. The total distance travelled and the magnitude of displacement are equal whenever the
object moves in one direction along a straight line and never turns back.
In Fig. 4.4 the athlete does turn back, so the two differ:

between t = 0 s and t = 16 s:

total distance travelled = OA + AB = 100 m + 60 m = 160 m

magnitude of displacement = OB = 40 m

160 m ≠ 40 m

But between t = 0 s and t = 10 s she runs only from O to A, without reversing:

total distance travelled = OA = 100 m

magnitude of displacement = OA = 100 m → equal

Why it happens: distance adds up every bit of path, treating forward and backward
alike. Displacement subtracts backward motion from forward motion. As long as
there is no backward motion to subtract, the two sums are identical. The moment
the object reverses, the return path adds to the distance but cancels part of the
displacement — so from then on the magnitude of displacement is always less than
the distance travelled.

Activity 4.1: Let us analyse — Page 51

Page 3 of 75

as e
Class 9 Science Chapter 4 Describing Motion Around Us
a g l AglaSem · NCERT Solutions

Section 4.1.2 Distance travelled and displacement
co m
e m.
m as
ACTIVITY

.co a g l
s m in Fig. 4.5, a ball is thrown vertically upwards from O. It moves up straight
eshown
gl aAs
a
Q1
till B and then falls back to O. Can this be considered a motion in a straight line?

co m
e m . ag
g l as
a

co m
em.
m l as
m .co a g
l a se
a g
m a s
m .co agl
l a se
a g

co m
m .
m as e
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se m
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a
se m
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m . a
ase
agl

co m
m .
m as e
.co a g l
se m
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a c
m .
m a s e
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g l as
a

co m
m .
m ase
.co


a g l Page 4 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

B
140 cm

120 cm

100 cm

80 cm C

60 cm

40 cm A

20 cm

0 cm O

Page 5 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Fig. 4.5, page 51 — a ball in vertical motion. The upward leg (left) and the downward leg
(right) are drawn as two separate dotted lines only for clarity; in reality the ball goes up
and falls back along the same line.

Yes. The whole journey — up from O to B and down again from B to O — happens along one
and the same vertical straight line.

Why it happens: motion in a straight line does not mean the object must keep
going one way. It only means the path is a straight line. Here the ball retraces the
very same line on the way down; Fig. 4.5 draws the up-leg and the down-leg as two
separate dotted lines only so that the marked positions A, B and C do not overlap on
the page. In reality both legs lie on top of each other.

Check it yourself: because the ball does turn back at B, this is a case where distance
travelled and magnitude of displacement will part company — exactly what Table 4.1
is designed to show.

Page 6 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Q2 For this motion, fill up the values in Table 4.1.

B
140 cm

120 cm

100 cm

80 cm C

60 cm

40 cm A

20 cm

0 cm O

Page 7 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Fig. 4.5, page 51 — a ball in vertical motion. The upward leg (left) and the downward leg
(right) are drawn as two separate dotted lines only for clarity; in reality the ball goes up
and falls back along the same line.

S. POSITION TOTAL DISTANCE DISPLACEMENT OF THE
NO. TRAVELLED BY THE BALL BALL FROM O TILL THAT
FROM O TILL THAT POSITION
POSITION

1. O 0 cm 0 cm

2. A 40 cm 40 cm in upward direction

3. B

4. C

5. O

Table 4.1, page 51 — the table as printed in the book, to be filled in.

Reading Fig. 4.5: O is at 0 cm, A at 40 cm, the highest point B at 140 cm, and C at 80 cm — but C
is marked on the downward leg, so the ball reaches C only after it has already come down 60 cm
from B.

S. POSITION TOTAL DISTANCE TRAVELLED BY DISPLACEMENT OF THE BALL
NO. THE BALL FROM O TILL THAT FROM O TILL THAT POSITION
POSITION

1. O 0 cm 0 cm

2. A 40 cm 40 cm in upward direction

3. B 140 cm 140 cm in upward direction

4. C 200 cm 80 cm in upward direction

5. O 280 cm 0 cm

Page 8 of 75

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Class 9 Science Chapter 4 Describing Motion Around Us
a g l AglaSem · NCERT Solutions

co m
em.
at C: distance = OB + BC = 140 cm + (140 cm − 80 cm) = 140 cm + 60 cm = 200 cm
m l as
.co
displacement = position of C − position of O = 80 cm − 0 cm = 80 cm upward
m a g
l a se
g
back at O: distance = OB + BO = 140 cm + 140 cm = 280 cm
a displacement = 0 cm − 0 cm = 0 cm

com
e m . ag
g l as
a
140 cm B
co m
se m.
m l a
m .co up: O →aAg → B
l a se = 140 cm
a g
80 cm C
a s
.com agl
m
ase
agl down: B → C → O
40 cm A
= 140 cm om .c
s e m
m a
e m . co agl
g l as 0 cm O O
a
se m
com g l a
m . a
ase
agl

co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


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Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Q3 Analyse the data filled in Table 4.1 and choose which of the following is true for
displacement: (i) It is never zero. (ii) Its magnitude can be greater than the total
distance travelled. (iii) Its magnitude is less than or equal to the total distance
travelled. (iv) Its magnitude is less than the total distance travelled in all cases.

Page 10 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

B
140 cm

120 cm

100 cm

80 cm C

60 cm

40 cm A

20 cm

0 cm O

Page 11 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Fig. 4.5, page 51 — a ball in vertical motion. The upward leg (left) and the downward leg
(right) are drawn as two separate dotted lines only for clarity; in reality the ball goes up
and falls back along the same line.

S. POSITION TOTAL DISTANCE DISPLACEMENT OF THE
NO. TRAVELLED BY THE BALL BALL FROM O TILL THAT
FROM O TILL THAT POSITION
POSITION

1. O 0 cm 0 cm

2. A 40 cm 40 cm in upward direction

3. B

4. C

5. O

Table 4.1, page 51 — the table as printed in the book, to be filled in.

The correct statement is (iii) Its magnitude is less than or equal to the total distance
travelled.
Test each option against the table:

(i) is false — at the last row the ball is back at O and the displacement is exactly 0 cm.
(ii) is false — at C the displacement is 80 cm while the distance travelled is 200 cm. The
displacement can never win, because it is only the straight-line gap between the two end
positions.
(iii) is true — at A both are 40 cm (equal); everywhere after the ball turns back, the
displacement is smaller.
(iv) is false — it says "in all cases", but at A they are equal, so "less than" does not hold in
every case.

Why it happens: the shortest route between two points is the straight line joining
them. The displacement is that straight-line gap; the distance travelled is the length
of the route actually taken. A route can never be shorter than the straight line, so
magnitude of displacement ≤ distance travelled, with equality only when the route is
the straight line — that is, when the object never turns back.

Page 12 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Pause and Ponder — Page 51
Section 4.1.2 Distance travelled and displacement

PAUSE AND PONDER

Q1 In the example of an athlete running back and forth on a straight track (Fig. 4.4),
when will the displacement of the athlete be zero? What will be the total distance
travelled in that case?

Total distance travelled

Displacement

t = 16 s
O B A

0m 20 m 40 m 60 m 80 m 100 m
t=0s t=4s t = 10 s

Fig. 4.4, page 50 — the athlete runs O → A (100 m) and back to B (60 m). Total distance
travelled 160 m; displacement 40 m from O.

The displacement is zero the moment she comes back to her starting point O — because
displacement is the net change in position, and her final position is then the same as her initial
position.

displacement = final position − initial position

=0m−0m=0m

If she runs from O all the way to A (100 m) and returns to O, then

total distance travelled = OA + AO

= 100 m + 100 m = 200 m

Page 13 of 75

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Class 9 Science Chapter 4 Describing Motion Around Us
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: the two quantities answer different questions. "How far did you

m l a se
walk?" adds every metre of the path — the return trip counts again. "Where are you
o gSo a closed trip
.c with where you began?" ignores the path entirely.
a
m
now, compared

l a se gives zero displacement and a non-zero distance.
ag
always

o m
c ag
Check it yourself: her displacement is zero for any round trip that ends at O — O to
.
m 80 m of distance, and O to A to O to A to O
s e
B and back gives 0 m displacement with
a of distance.
gives 0 m displacement with 400lm
ag

. com
m a s emyour answer. (i)
Fuel used up in a vehicle depends on which of the following? Justify
codistance travelled (ii) Displacement gl
Q2

. a
em
Total

l a s
agANSWER
(i) Total distance travelled.
m a s
m.co agl
l a se
Why it happens: the engine has to keep pushing the vehicle against friction and air
a g
resistance over every single metre of the road it actually covers. That work — and
therefore the fuel burnt — piles up along the whole path. Displacement carries no

com
information about the path at all.
m .
m as e
.co a g l
emaway, and drives back home:
A quick test settles it. Suppose an auto-rickshaw picks you up from home, drops you at school 4

a s
gl
km

a
se m
com a
total distance travelled = 4 km + 4 km = 8 km
. a g l
m
ase
displacement = 0 km

a gl
If fuel depended on displacement, the return journey would have to be free — and the tank

co m
would be as full at the end as at the start. It clearly is not. The fuel gauge and the odometer

m .
e
both track distance, not displacement.
m l as
.co a g
a s em Did you know? This is exactly why an auto or taxi meter runs on the odometer

agl reading (distance), never on how far you ended up from where you started.
.c
s e m
m a
e m . co agl
g l as
a

co m
m .
m ase
.co


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Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Q3 A ball rolls down an inclined track as shown in Fig. 4.6. Is its motion, a straight line
motion? Assuming the starting point of the ball (O) to be the origin, can its motion
from O to D be depicted using a horizontal line as shown in Fig. 4.3? Are the values
of total distance travelled and magnitude of displacement from O equal or different
at positions A, B, C and D?

40 cm
10 cm
O 20 cm
30 cm
A
B
C
D

Fig. 4.6, page 51 — a ball rolling down an inclined track. OA = 40 cm, AB = 10 cm, BC = 20
cm, CD = 30 cm.

O B A
− +

−20 m 0m 20 m 40 m 60 m 80 m 100 m

Fig. 4.3, page 50 — the reference point O and the athlete's positions B and A at two
instants, marked on a straight line with distances.

Yes, it is straight-line motion. The track in Fig. 4.6 is a straight (though sloping) line, and the
ball stays on it throughout — so the path is a straight line.
Yes, it can be shown on a line like Fig. 4.3. The motion is one-dimensional: one number (the
distance along the track from O) fixes the ball's position completely. Drawing that number line
horizontally on paper is only a convenience — the line stands for the track itself, and the
positive direction along it is "down the slope from O towards D", not "horizontally to the right".
Reading the markings of Fig. 4.6 along the track: OA = 40 cm, AB = 10 cm, BC = 20 cm, CD = 30
cm. So the positions measured from O are

Page 15 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

POSITION TOTAL DISTANCE TRAVELLED MAGNITUDE OF DISPLACEMENT EQUAL?
FROM O FROM O

A 40 cm 40 cm Yes

B 40 + 10 = 50 cm 50 cm Yes

C 50 + 20 = 70 cm 70 cm Yes

D 70 + 30 = 100 cm 100 cm Yes

So at A, B, C and D the two are equal.

Why it happens: the ball moves down the incline in one direction only and never
rolls back up. With no reversal, the path is the straight line joining the two positions,
so its length equals the straight-line gap. The direction of the displacement, however,
is along the incline — pointing down the slope — and not horizontal.

Pause and Ponder — Page 53
Section 4.1.3 Average speed and average velocity

PAUSE AND PONDER

Q4 During a family road trip, you drive 200 km north in three hours. Afterwards, you
drive 200 km south in two hours. Find the average speed and average velocity for
your entire trip.

Handle the two quantities separately — one uses the path, the other only the end points.

total distance travelled = 200 km + 200 km = 400 km

total time interval = 3 h + 2 h = 5 h

average speed = total distance travelled / time interval

= 400 km / 5 h = 80 km h⁻¹

For the velocity, take north as positive:

Page 16 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

displacement = (+200 km) + (−200 km) = 0 km

average velocity = displacement / time interval

= 0 km / 5 h = 0 km h⁻¹

Why it happens: you finish exactly where you started, so the net change in position
is zero — and average velocity, being displacement over time, must be zero however
fast you drove. The average speed is not zero because the odometer counted all 400
km. Note also that the average speed is not the average of the two leg speeds: leg
speeds are 200/3 ≈ 66.7 km h⁻¹ and 200/2 = 100 km h⁻¹, whose plain average is 83.3
km h⁻¹, not 80 km h⁻¹. Always divide total distance by total time.

In SI units: 80 km h⁻¹ = 80 × 1000 m / 3600 s = 22.2 m s⁻¹.

Q5 Under what condition(s) is the (i) magnitude of average velocity of an object equal
to its average speed? (ii) magnitude of average velocity of an object zero while its
average speed is not zero?

(i) When the object moves along a straight line in one direction only, without ever turning back,
during that whole time interval.

|average velocity| = |displacement| / t and average speed = distance / t

these are equal exactly when |displacement| = distance travelled

which happens only when the path is a straight line covered in one direction

(ii) When the object returns to its starting position at the end of the interval, so the
displacement is zero, while the path length is not.

displacement = 0 → average velocity = 0

distance travelled ≠ 0 → average speed ≠ 0

Page 17 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Examples: Sarang in Example 4.2 swims 50 m up and back in 50 s — average speed 1 m s⁻¹,
average velocity 0 m s⁻¹. So does anything completing a full revolution of a circle, or a train that
returns to the same platform.

Why it happens: the numerator is the only difference between the two definitions
— distance for speed, displacement for velocity. Case (i) is the situation where those
two numerators coincide; case (ii) is the extreme where the displacement numerator
collapses to zero while the distance numerator keeps growing.

Activity 4.2: Let us calculate — Page 55
Section 4.1.4 Average acceleration

ACTIVITY

Q1 The magnitude of average acceleration of cars is generally specified as the time
taken by the car to go from 0 km h⁻¹ to 100 km h⁻¹. Look it up on the internet and
find this time for various cars, and record those in Table 4.2.

CAR TIME INTERVAL DURING WHICH MAGNITUDE OF AVERAGE
TYPE THE SPEED GOES FROM 0 TO 100 KM ACCELERATION (M S⁻²)
H⁻¹ (S)

Table 4.2, page 55 — the magnitude of average acceleration in a time interval (printed
blank in the book, for you to fill in).

First convert the change in speed into SI units, because the acceleration must come out in m s⁻².

change in speed = 100 km h⁻¹ − 0 km h⁻¹ = 100 km h⁻¹

100 km h⁻¹ = 100 × 1000 m / 3600 s = 27.8 m s⁻¹

Now look up the "0–100 km h⁻¹" time quoted by the manufacturer or a road test for each car
and fill the middle column. Typical published values look like this — treat them as a model; the
whole point of the activity is that you find and record real figures yourself.

Page 18 of 75

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Class 9 Science Chapter 4 Describing Motion Around Us
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m
.co
CAR TYPE TIME INTERVAL DURING WHICH THE MAGNITUDE OF AVERAGE

se m
com a
SPEED GOES FROM 0 TO 100 KM H⁻¹ (S) ACCELERATION (M S⁻²)

. ag l
Small
a s em 12.0 2.3

agl
hatchback

Mid-size 10.0 2.8

com
. ag
sedan

e m
SUV (turbo- 8.5
g l as 3.3
petrol) a
Electric 7.0 4.0
co m
em.
hatchback

m l as
m
Sports car
.co 3.5 a g
se
7.9

g l a
a Tip: keep the third column to two significant figures. The published 0–100 time is

m a s
decimals would be false precision. .co agl
itself only quoted to about a tenth of a second, so quoting the acceleration to three

a s em
agl

co m
.
Q2 Calculate the magnitude of average acceleration for each car.

se m
o m g l a
m .c a
se the formula, then substitute — the same three lines for every row of the table.

g l aState
a
se m
com
average acceleration = (final velocity − initial velocity) / time interval [Eq. 4.3b]
g l a
m . a
ase
|a| = (27.8 m s⁻¹ − 0 m s⁻¹) / t

agl
Worked out for the sample rows:

co m
m .
m as e
.co l
t = 12.0 s → |a| = 27.8 / 12.0 = 2.3 m s⁻²
a g
s e m t = 10.0 s → |a| = 27.8 / 10.0 = 2.8 m s⁻²
agla
.c
t = 8.5 s → |a| = 27.8 / 8.5 = 3.3 m s⁻²

s e m
om a
agl
t = 7.0 s → |a| = 27.8 / 7.0 = 4.0 m s⁻²
. c
s e m m s⁻ ²
t = 3.5 s → |a| = 27.8 / 3.5 = 7.9
a
agl

com
m .
m ase
.co


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Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Why it happens: every car reaches the same final speed, so the numerator 27.8 m
s⁻¹ is fixed and the acceleration is simply inversely proportional to the time. A car
that halves its 0–100 time doubles its average acceleration. Compare these with g =
9.8 m s⁻²: even a sports car accelerates forward at less than the rate at which a
dropped stone gains downward speed.

Check it yourself: the figure you get is an average over the whole run. The real
acceleration is largest in the lower gears and falls off at high speed, where air
resistance grows — so the car is never accelerating at exactly this value.

Activity 4.3: Let us plot a graph — Page 57
Section 4.2.1 Plotting graph

ACTIVITY

Q1 Refer to Table 4.3. We need to decide which quantity (time or position) to be shown
along each axis.

TIME 0S 1S 2S 3S 4S 5S 6S

Position 0m 20 m 40 m 60 m 80 m 100 m 120 m

Table 4.3, page 57 — positions of the vehicle at different instants of time.

Put time along the X-axis and position along the Y-axis.

Why it happens: time is the quantity we control and read off steadily — it marches
on whatever the vehicle does. Position is the quantity that responds to it. The
convention in every graph is that the controlling (independent) quantity goes on the
X-axis and the responding (dependent) quantity on the Y-axis. That way the slope,
rise ÷ run, automatically reads as "change in position ÷ change in time" — which is
the velocity, a quantity we actually want. Swap the axes and the slope would come
out as time per metre, which means nothing useful.

Page 20 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Tip: the same rule gives the velocity–time graph its meaning later: with time on X
and velocity on Y, the slope is acceleration and the area is displacement.

Q2 Determine a suitable scale for each quantity to represent it on the graph paper.

TIME 0S 1S 2S 3S 4S 5S 6S

Position 0m 20 m 40 m 60 m 80 m 100 m 120 m

Table 4.3, page 57 — positions of the vehicle at different instants of time.

A good scale is the one that spreads the data over most of the sheet while keeping the numbers
easy to plot. For Table 4.3 the book's choice is

X-axis: 5 divisions = 1 s (time runs 0 s to 6 s)

Y-axis: 5 divisions = 20 m (position runs 0 m to 120 m)

Check what that uses up: 6 s × 5 divisions = 30 divisions across, and 120 m ÷ 20 m × 5 = 30
divisions up — a comfortable square block on an ordinary sheet.

Why it happens: two conditions decide a scale. (a) The largest value must fit on the
paper. (b) Each small division must stand for a round number (1, 2, 5, 10, 20 …) so
that intermediate points can be read without arithmetic. A scale like "5 divisions = 17
m" would fit but nobody could plot 60 m on it. And a scale far too large, say 5
divisions = 100 m, would squeeze all six points into a corner and the slope would be
impossible to measure accurately.

Check it yourself: the scale changes how the line looks, never what it means.
Steepness on paper depends on the scales you chose; the velocity you calculate from
it does not.

Page 21 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Q3 Once all points are plotted, connect them to create the position-time graph for the
vehicle's motion (Fig. 4.11c). It is a straight line for the data given in Table 4.3.

TIME 0S 1S 2S 3S 4S 5S 6S

Position 0m 20 m 40 m 60 m 80 m 100 m 120 m

Table 4.3, page 57 — positions of the vehicle at different instants of time.

Plotting (0 s, 0 m), (1 s, 20 m), (2 s, 40 m) … (6 s, 120 m) and joining them gives a single straight
line through the origin.

Position (m)
120

100

80

60

40

20

0 1 2 3 4 5 6
Time (s)
Position–time graph for Table 4.3. Equal position gains of 20 m in every 1 s make the points fall on
one straight line.

Page 22 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Why it happens: in each successive 1 s the vehicle's position increases by exactly 20
m. Equal rises for equal runs is precisely the geometric condition for a straight line.
Its slope is the velocity:

v = (120 m − 0 m) / (6 s − 0 s) = 20 m s⁻¹, constant throughout

So a straight-line position–time graph means constant velocity, i.e. uniform motion.

Tip: this graph is not a picture of the road. It does not say the vehicle went "up" — it
says its distance from the origin grew steadily with time.

In-text Questions — Page 59
Section 4.2.2 Position-time graphs

Q1 What does the shape of the position-time graph indicate about the nature of
motion?

The shape tells you at once whether the velocity is constant or changing, because the slope of a
position–time graph is the velocity.

SHAPE OF THE POSITION– WHAT IT MEANS REASON
TIME GRAPH

Straight line, sloping Constant velocity (uniform Equal displacements in equal times, so
motion) the slope never changes

Straight line parallel to the time Object at rest Position does not change with time, so
axis the slope is zero

Curve getting steeper Velocity increasing — Larger displacements in successive equal
accelerated motion time intervals

Curve getting flatter Velocity decreasing Smaller displacements in successive
equal time intervals

The book's own two examples make it concrete. In Fig. 4.13a the object covers 20 m between 2 s
and 3 s and again 20 m between 5 s and 6 s — equal displacements in equal times, so the graph
is straight and the velocity is constant. In Fig. 4.13b the displacement between 4 s and 6 s is
smaller than between 10 s and 12 s — the graph curves upward and the velocity is increasing.

Page 23 of 75

as e
Class 9 Science Chapter 4 Describing Motion Around Us
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: velocity is defined as change in position ÷ change in time, and on

m l a se
this graph "change in position" is the rise and "change in time" is the run. So velocity
o
.c A shape with fixed steepness must mean fixed a g
m
is the steepness. velocity; a shape

l a se steepness changes must mean changing velocity, which is exactly
g
whose
aacceleration.

co m
e m . ag
g l as
a
Q2 Which physical quantities can be obtained from a position-time graph?

com
m.

m
Three things can be pulled straight off it:
o l a se
g above that time.
.c of the object at any instant — read the Y-value directly
a
m
se displacement between any two instants — subtract the two Y-values, s₂ − s₁.
1. The position
2.aThe
l
g
a 3. The average velocity between those instants — the slope of the line joining the two
points.
m a s
m .co agl
l a se
a g
v = BC / CA = (s₂ − s₁) / (t₂ − t₁) [Eq. 4.2a]

using Fig. 4.14: v = (80 m − 40 m) / (4 s − 2 s) = 40 m / 2 s = 20 m s⁻¹

. com
m line),
You can also tell, without any calculation, whether the motion is uniforme(straight
s
m a
glcomparing two lines on the
co (curve) or at rest (line parallel to the time axis) — andaby
.
em axes, which object is faster (Example 4.7: the steeper line has the larger velocity).
accelerated

a s
gl
same

a
Why it happens: the graph stores position as a function of time, and every one of
se m
com g l a
. a
these quantities is built out of position and time alone. What it cannot give you
m
ase
directly is the acceleration — for that you would need how the slope itself is

agl
changing, which is a further step you will meet in higher grades.

co m
m .
m 4.4: Let us calculate — Pages 59–60 as e
.co
Activity
a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

com
m .
m ase
.co


a g l Page 24 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Section 4.2.2 Position-time graphs

ACTIVITY

Q1 In the position-time graph we plotted (Fig. 4.11c), consider a part (say, AB) of the
graph as shown in Fig. 4.14. From A, draw a line parallel to X-axis and another line
parallel to Y-axis. Repeat the same from B.

120

100
s₂ B
Position (m)

80

60
A C
40
s₁
20
t₁ t₂

0 1 2 3 4 5 6
Time (s)

Fig. 4.14, page 59 — calculating velocity from a position–time graph. A is (t₁ = 2 s, s₁ = 40
m) and B is (t₂ = 4 s, s₂ = 80 m); BC = s₂ − s₁ and CA = t₂ − t₁.

Take A at t₁ = 2 s (position s₁ = 40 m) and B at t₂ = 4 s (position s₂ = 80 m) on the straight line of
Fig. 4.11c.
From A, the line parallel to the Y-axis drops to t₁ on the time axis; the line parallel to the X-axis
runs across to s₁ on the position axis. Doing the same from B marks off t₂ and s₂.

Page 25 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Position (m)

B
80 m

BC = 40 m
CA = 2 s
40 m
A C

t₁ = 2 s t₂ = 4 s
Time (s)
Dropping perpendiculars from A and B on the position–time line marks off the time interval CA and
the change in position BC.

Tip: the two dashed lines from each point are just a way of reading the coordinates
of A and B off the axes. They do not represent any motion.

Q2 Extend the horizontal line from A and a triangle ABC is formed. What do the sides
BC and CA of the triangle represent?

The right-angled triangle ABC has the graph line AB as its hypotenuse. Its two perpendicular
sides are:

CA — the horizontal side — represents the change in time, t₂ − t₁.
BC — the vertical side — represents the change in position, s₂ − s₁, which is the displacement
between those two instants.

CA = t₂ − t₁ = 4 s − 2 s = 2 s
BC = s₂ − s₁ = 80 m − 40 m = 40 m

Page 26 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Why it happens: CA lies along the direction in which the X-axis measures time, so its
length can only be a time interval; BC lies along the direction in which the Y-axis
measures position, so its length can only be a change in position. This is why the
triangle is worth drawing at all — it converts the abstract "slope" into two sides you
can actually measure with the graph's own scales.

Q3 As per Eq. (4.2a), by dividing the change in position (BC) by the change in time (CA),
you get the average velocity.

Yes — dividing the two sides of the triangle reproduces the definition of average velocity exactly.

average velocity = change in position / time interval [Eq. 4.2a]

here change in position = BC and time interval = CA

v = BC / CA = (s₂ − s₁) / (t₂ − t₁)

Why it happens: BC ÷ CA is, geometrically, the slope of the line AB — how steeply it
rises. Physically the same ratio is displacement ÷ time, which is velocity. So "slope of
a position–time graph" and "velocity" are not two facts to remember; they are the
same ratio read in two languages. In general, the slope of any graph gives the rate
of change of the Y-quantity with respect to the X-quantity.

Page 27 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Q4 By extracting values of time t₁ and t₂, and distances s₁ and s₂ from the graph, the
magnitude of average velocity can be calculated.

120

100
s₂ B
Position (m)

80

60
A C
40
s₁
20
t₁ t₂

0 1 2 3 4 5 6
Time (s)

Fig. 4.14, page 59 — calculating velocity from a position–time graph. A is (t₁ = 2 s, s₁ = 40
m) and B is (t₂ = 4 s, s₂ = 80 m); BC = s₂ − s₁ and CA = t₂ − t₁.

Substituting the four values read off Fig. 4.14:

v = (s₂ − s₁) / (t₂ − t₁)

= (80 m − 40 m) / (4 s − 2 s)

= 40 m / 2 s

= 20 m s⁻¹

Because the graph is a straight line, every other pair of points gives the same answer. Check
with t₁ = 0 s, t₂ = 6 s:

v = (120 m − 0 m) / (6 s − 0 s) = 120 m / 6 s = 20 m s⁻¹ ✓

Page 28 of 75

ase
Class 9 Science Chapter 4 Describing Motion Around Us
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: a straight line has one slope everywhere. That is exactly what

m l a se
"constant velocity" means, and it is why the choice of the segment AB does not
o g segments
.c For a curved position–time graph it would matter: different
a
m
matter here.

l a segive different average velocities, because the velocity itself is changing.
ag
would

o m
c ag
Check it yourself: carry the units through the division, as above. Metres divided by
seconds gives m s⁻¹ automatically — am .
divided the wrong way round. la s e useful check that you have not accidentally

ag

co m
se m.
c o m
In-text Questions — Pages 61–62
g l a
Section 4.2.3.Velocity-time a
se m graphs

l a
ag Q1 What does the shape of the velocity-time graph indicate about the nature of
m a s
.co agl
motion?

se m
g l a
a

Here the slope is the acceleration, so the shape reports on how the velocity is changing.

co m
m .
e
SHAPE OF THE WHAT IT MEANS BOOK'S EXAMPLE

m l as
.co
VELOCITY–TIME GRAPH

a g
s m line parallel to the time
eStraight
a
gl axis
Fig. 4.17a — car at a
Velocity constant, acceleration zero

a steady 20 m s⁻¹

se m
com and along the velocity l a
Straight line sloping upward Velocity increasing at a constant rate; Fig. 4.17b — 0 to 15 m

. a g
emdecreasing at a constant rate;
acceleration constant s⁻¹ in 30 s

l a s
agacceleration
Straight line sloping downward Velocity Fig. 4.17c — 15 m s⁻¹
constant and opposite to the down to 0 in 30 s

omtreated in this
velocity

. cNot
e m
as
Curved line Acceleration itself is changing
m l
.co a g chapter

a s emTable 4.5 shows why Fig. 4.17b is straight: the velocity rises by exactly 2.5 m s⁻¹ in every 5 s
agl
.c
m
interval — equal changes in equal times.

m a s e
. c o
a = (2.5 m s⁻¹ − 0 m s⁻¹) / (5 s −m0 s) = 0.5 m s⁻² agl
l a se
g / (10 s − 5 s) = 0.5 m s⁻² … the same throughout
a = (5.0 m s⁻¹ − 2.5 mas⁻¹)

co m
m .
m as e
.co


a g l Page 29 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Why it happens: acceleration is change in velocity ÷ change in time, which on these
axes is rise ÷ run — the slope. A fixed slope therefore means a fixed acceleration,
and a zero slope means no acceleration at all, however large the velocity may be.

Q2 Which physical quantities can be obtained from a velocity-time graph?

Three, and the third one is the reason this graph is so useful:

1. The velocity at any instant — read the Y-value directly.
2. The average acceleration — the slope of the line between two points.
3. The displacement — the area enclosed between the line and the time axis for that interval.

slope: a = BC / CA = (v − u) / (t₂ − t₁)

from Fig. 4.17d between 10 s and 20 s:

a = (10 m s⁻¹ − 5 m s⁻¹) / (20 s − 10 s) = 5 m s⁻¹ / 10 s = 0.5 m s⁻²

area: displacement between 0 s and 6 s in Fig. 4.18a

= area of rectangle OABC = OA × OC

= 20 m s⁻¹ × 6 s = 120 m

Why it happens: the area is a product of a height (velocity, in m s⁻¹) and a width
(time, in s), and m s⁻¹ × s = m — a length. That is not a coincidence: for constant
velocity, displacement = velocity × time is exactly the area of the rectangle. For a
sloping line the same idea holds; you just split the shape into a rectangle and a
triangle.

Q3 Can you calculate some other physical quantity from the velocity-time graph?

Yes — the displacement, from the area enclosed between the graph line and the time axis.
For the accelerating car of Fig. 4.18b, the displacement between 10 s and 20 s is the area of the
shape ABDE, which splits into a rectangle and a triangle:

Page 30 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

displacement = area of ABDE

= area of rectangle ACDE + area of triangle ABC

= (CD × DE) + (½ × CA × BC)

= (5 m s⁻¹ × 10 s) + (½ × 10 s × 5 m s⁻¹)

= 50 m + 25 m = 75 m

Velocity (m s⁻¹)

B
10

25 m
A
5
C
50 m

E D
10 s 20 s

Displacement between 10 s and 20 s = rectangle (50 m) + triangle (25 m) = 75 m.

Why it happens: the rectangle is the distance the car would have covered had it
kept its starting velocity of 5 m s⁻¹ for the full 10 s. The triangle is the extra distance it
gains because it is speeding up. Adding them is just adding "what it would have
done" to "what the acceleration added".

Bridging Science and Society — Page 65

Page 31 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Section 4.3 Kinematic equations

BRIDGING SCIENCE AND SOCIETY

Q1 Can you now understand why it is important to maintain a safe distance from the
vehicle moving ahead of your vehicle (Fig. 4.20) and how this distance needs to be
adjusted given your initial velocity?

safe distance

direction of motion

Fig. 4.20, page 65 — safe distance between two moving vehicles (redrawn as a labelled
sketch from the illustration in the textbook).

Yes. A vehicle cannot stop where the driver decides to stop — it stops where physics allows, and
that place is set by u², not by u.

stopping distance = reaction distance + braking distance

= u tr + u² / (2|a|) [the second term from v² = u² + 2as with v = 0]

Using the chapter's own braking figure of |a| = 4 m s⁻² and a reaction time tr = 1 s:

INITIAL REACTION BRAKING DISTANCE TOTAL STOPPING
VELOCITY U DISTANCE U TR U²/2|A| DISTANCE

36 km h⁻¹ = 10 m s⁻¹ 10 m 12.5 m 22.5 m

54 km h⁻¹ = 15 m s⁻¹ 15 m 28.1 m 43.1 m

108 km h⁻¹ = 30 m s⁻¹ 30 m 112.5 m 142.5 m

Going from 54 to 108 km h⁻¹ doubles the speed but makes the stopping distance more than
three times longer — because the braking part alone becomes four times longer.

Page 32 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Why it happens: during braking the vehicle loses kinetic energy at a roughly steady
rate set by the friction the tyres can supply, and the speed enters that balance as a
square. The reaction part is different in nature: it is a stretch of road covered at full
speed with no braking at all, so it simply grows in proportion to u. This is why a fixed
"keep 20 m behind" rule is useless — the safe gap must grow with speed, which is
why drivers are taught a time gap (2–3 s) instead.

Did you know? Anything that lowers |a| — a wet or gravelly road, worn tyres, a
heavily loaded truck — stretches the braking term inversely. Halve the grip and every
braking distance in the table doubles. Vehicle-to-vehicle (V2V) technology, now being
developed in India and elsewhere, attacks the other term instead: it lets the vehicle
ahead warn you electronically, cutting the reaction time far below a human's.

In-text Questions — Page 66
Section 4.4.1 Uniform circular motion

Q1 What is the distance travelled by the child? What is their displacement from their
original position?

The child on the merry-go-round moves from A to B to C along the rim (Fig. 4.22).

Distance travelled = the length of the curved arc ABC actually followed along the circle.
Displacement = the straight line AC, joining the starting and finishing positions, directed
from A towards C.

The two are not equal: the arc bulges outward while the chord cuts straight across, so arc ABC >
chord AC.

Why it happens: on a circular path the object is turning at every instant, so it is
never travelling straight towards its destination. Every bit of the arc is longer than
the straight-line progress it makes. This gap between path length and net change in
position is the clearest reason why we need two separate quantities — distance and
displacement — rather than one.

Page 33 of 75

as e
Class 9 Science Chapter 4 Describing Motion Around Us
a g l AglaSem · NCERT Solutions

co m
m.
What is the distance travelled by the child in making one revolution (going round
e
Q2

m l as
.co
the circle once)?

a g
se m
g l a
a

One full revolution means going all the way round the rim once, so the distance travelled is the

m
circumference of the circular path.
. co ag
e m
g l as
a
distance in one revolution = 2πR

where R is the radius of the circular path

. com
m a s em at exactly
gl
The displacement, on the other hand, is zero, because the child ends the revolution
the same c
. o from which it began. a
emwhy the two averages part company so dramatically over one revolution. If the
position

l a s
agrevolution
That is
takes time T,

m a s
em
average speed, vav = 2πR / T [Eq. 4.5] .co agl
a s
agl/ T = 0 / T = 0
average velocity = displacement

co m
m .
e
Check it yourself: for a merry-go-round of radius R = 2 m turning once in T = 8 s, vav
m l as
.co g
= 2 × 3.14 × 2 m / 8 s = 12.6 m / 8 s = 1.6 m s⁻¹, while the average velocity over that

em a
s
revolution is 0 m s⁻¹.
l a
ag
se m
comthe speed is constant but what about the g l a
.
In case of uniform circular motion,
a
em Is it changing?
Q3

a s
agl
direction of velocity at an instant?

. com
Yes, the direction changes continuously — at every single instant. The velocity at a point is
mmotion, and the tangent
directed along the tangent to the circle at that point, in the directioneof
s
m
co as the object goes round. gl a
. a
em
turns

l a s The book's argument with the rectangular and hexagonal tracks (Fig. 4.23) shows why. On a
ag c
.
rectangle the runner changes direction 4 times per lap; on a hexagon, 6 times. Increase the
number of sides and the turns become more frequent and each one smaller. In the limit, the
s e m
m a
. co agl
track becomes a circle — each side shrinks to a point, and the turning never stops.

e m
g l as
a

co m
m .
m ase
.co


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Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Why it happens: velocity is not a number, it is a magnitude and a direction. A
change in either one is a change in velocity, and a changing velocity means non-zero
acceleration. So uniform circular motion is accelerated motion, even though the
speedometer reading would never move. We usually say a vehicle is "accelerating"
only when its speed changes, and so we miss the acceleration that a car actually has
while taking a circular turn at a steady speed.

Check it yourself: Activity 4.5 makes the tangent visible. Lift the ring while the
marble is circling inside it, and the marble shoots off in a straight line — along the
tangent at the instant it was freed, which is the direction its velocity had at that
moment.

Activity 4.5: Let us investigate — Page 67
Section 4.4.1 Uniform circular motion

ACTIVITY

Q1 Take a ring, such as an adhesive tape ring and one marble.

Use a stiff, fairly wide ring — the empty cardboard core of an adhesive-tape roll is ideal — and
one smooth glass marble that rolls freely inside it.

Tip: a wider ring gives a longer circular path, so the marble completes fewer
revolutions per second and you can follow its direction of motion by eye. A ring with
a smooth inner wall matters too: a rough wall keeps nudging the marble and the
"circular" motion becomes ragged.

Page 35 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Q2 Place the ring flat on a smooth surface and throw the marble inside the ring in a
way that it rotates along the inner boundary of the ring (Fig. 4.24).

marble

adhesive-tape ring

smooth surface

Fig. 4.24, page 67 — a marble moving inside a ring; redrawn here as a labelled top-view
sketch (the textbook prints a photograph).

Rest the ring flat on a smooth table or floor and flick the marble in along the inner wall, not
towards the centre, so that it hugs the boundary and circles round.

Why it happens: left to itself the marble would roll in a straight line. It goes round
only because the inner wall of the ring keeps pushing it inwards, continuously
bending its path. That inward push from the wall is what turns straight-line motion
into circular motion — and it is also what will be missing the moment you lift the
ring.

Tip: a smooth surface matters. On a rough floor friction slows the marble quickly,
the speed stops being constant, and the motion is no longer uniform circular motion.

Page 36 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Q3 Predict what will happen if you lift the ring while the marble is moving.

Prediction: the marble will stop curving and will roll away in a straight line — along the tangent
to the circle at the point where it happened to be when the ring left it.
It will not keep circling, and it will not fly outward along a radius from the centre. Write your
prediction down before you test it.

Why it happens: at every instant the marble's velocity already points along the
tangent — that is the direction it is actually moving. The wall was only bending that
velocity round. Remove the wall, and nothing is left to change the direction, so the
marble simply carries on the way it was already going at that instant.

Q4 Now, after one or two complete revolutions of the marble, pick up the ring without
disturbing the motion of the marble. What do you observe? Does the marble
continue moving in a circular motion? Or does it move in some other manner?

Observation: the marble immediately leaves the circle and rolls off in a straight line. It does
not continue in circular motion.
The straight line it takes is the tangent to the circle at the exact point where the marble was
when the ring was lifted — so if you repeat the activity and lift the ring at a different point, the
marble goes off in a different direction.

Why it happens: once the marble is released it continues to move in the direction it
was already moving at that instant, and with no wall pushing it sideways there is
nothing to change that direction. This confirms two things at once: (a) the velocity in
circular motion is along the tangent, and (b) circular motion needs something acting
continuously to keep bending the path. You will learn the reason behind (b) in a later
chapter.

Q5 Repeat the activity multiple times to confirm the result.

Do it at least four or five times, lifting the ring at a clearly different point of the circle each time,
and mark or note the direction the marble takes away.

Page 37 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Every trial should give the same finding: a straight-line escape, always along the tangent at the
release point.

Why it happens: a single trial could be luck — the marble might have been nudged,
or the surface might have sloped. Repeating with the release point deliberately
changed tests the rule, not one instance of it. If the marble left along a radius, or
kept curving, the tangent explanation would be wrong. It never does, which is what
makes the conclusion trustworthy.

Check it yourself: if the marble slows visibly during the revolutions, your surface is
not smooth enough — move to a glass tabletop or a well-polished floor, so that the
circular motion stays close to uniform.

Revise, Reflect, Refine — Pages 68–70
End-of-chapter questions

REVISE, REFLECT, REFINE

Q1 My father went to a shop from home which is located at a distance of 250 m on a
straight road. On reaching there, he discovered that he forgot to carry a cloth bag.
He came home to take it, went to the shop again, bought provisions and came back
home. How much was the total distance travelled by him? What was his
displacement from home?

Break the trip into its four legs. Each leg is 250 m long.

LEG JOURNEY DISTANCE

1 home → shop 250 m

2 shop → home (for the bag) 250 m

3 home → shop (again) 250 m

4 shop → home (with provisions) 250 m

total distance travelled = 250 m + 250 m + 250 m + 250 m

= 4 × 250 m = 1000 m = 1 km

Page 38 of 75

ase
Class 9 Science Chapter 4 Describing Motion Around Us
a g l AglaSem · NCERT Solutions

For the displacement, take home as the origin and the direction of the shop as positive:
co m
e m.
m l as
m .co
displacement = final position − initial position
a g
l a se
g
=0m−0m=0m
a

. com
Why it happens: he ends the whole outing standing exactly where he started, so the
ag
a s em how many times he walked the road. The
net change in position is nil — no matter

agl metre of walking, and the forgotten bag cost him
distance, in contrast, counts every
an extra 500 m of it.

co m
se m.
o m l a
ga school building to
Q2 m
.c a
se collect a book and then comes down to their classroom on the second floor. If the
A student runs from the ground floor to the fourth floor of
l a
ag height of each floor is 3 m, find: (i) the total vertical distance travelled, and (ii) their
displacement from the starting point.
m a s
m.co agl
l a se
g

a
Height of one floor = 3 m. Measure heights above the ground floor, taking upward as positive.

co m
m .
e
ground floor → fourth floor: rise of 4 floors = 4 × 3 m = 12 m (upward)
m l as
m .co a g
fourth floor → second floor: fall of 2 floors = 2 × 3 m = 6 m (downward)

l a se
ag (i) Total vertical distance travelled

se m
com g l a
m . a
ase
= 12 m + 6 m = 18 m

a gl
(ii) Displacement from the starting point

co m
m .
height of second floor above ground floor = 2 × 3 m = 6 m
as e
com
.displacement a g l
se m = 6 m − 0 m = 6 m, vertically upward

g l a
a c
m .
Why it happens: the downward 6 m is real climbing — the legs feel it — so it adds to
s e
. com
the distance. But it cancels part of the upward climb when we ask only "how much
a gla
a s em at the start?" Hence 12 + 6 = 18 m for distance, but 12
higher is the student now than

agl
− 6 = 6 m for displacement.

co m
m .
m ase
.co


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Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Tip: the numbers are floor gaps, not floor numbers. From the ground floor to the
fourth floor there are 4 gaps, not 5.

Q3 A girl is riding her scooter and finds that its speedometer reading is constant. Is it
possible for her scooter to be accelerating and if so, how?

Yes, it certainly can be accelerating — if she is changing direction, for example going round a
bend or a roundabout.

Why it happens: the speedometer reports only the magnitude of the velocity.
Velocity, however, is magnitude and direction, and acceleration is the rate of change
of velocity. So a change in direction alone is enough to give a non-zero acceleration
even while the needle sits perfectly still.

acceleration ≠ 0 if magnitude of velocity changes, or direction of velocity changes, or

both

This is exactly the case of uniform circular motion in Section 4.4: constant speed,
continuously turning velocity, and therefore accelerated motion.

The only way her scooter can have zero acceleration is if it moves along a straight road with
that constant speedometer reading.

Check it yourself: you feel this acceleration as the sideways push while turning a
corner, even though the engine note and the speedometer do not change at all.

Q4 A car starts from rest and its velocity reaches 24 m s⁻¹ in 6 s. Find the average
acceleration and the distance travelled in these 6 s.

Given: u = 0 m s⁻¹ (starts from rest), v = 24 m s⁻¹, t = 6 s

Average acceleration — use Eq. (4.3b):

Page 40 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

a = (v − u) / t

= (24 m s⁻¹ − 0 m s⁻¹) / 6 s

= 24 m s⁻¹ / 6 s = 4 m s⁻², in the direction of motion

Distance travelled — use Eq. (4.4b):

s = ut + ½at²

= (0 m s⁻¹ × 6 s) + (½ × 4 m s⁻² × (6 s)²)
= 0 + ½ × 4 × 36 m

= 72 m

Check it yourself: for constant acceleration the average velocity is simply (u + v)/2 =
(0 + 24)/2 = 12 m s⁻¹, so s = 12 m s⁻¹ × 6 s = 72 m ✓. Equally, from Eq. (4.4c): v² = u² +
2as → 576 = 0 + 8s → s = 72 m ✓. Three routes, one answer.

Why it happens: the car is speeding up all through, so it covers less ground than a
car doing a steady 24 m s⁻¹ (which would manage 144 m) but more than one
crawling near zero. Exactly halfway, in fact — because with constant acceleration the
velocity rises linearly, so the average velocity is the plain mean of the start and end
values.

Q5 A motorbike moving with initial velocity 28 m s⁻¹ and constant acceleration stops
after travelling 98 m. Find the acceleration of the motorbike and the time taken to
come to a stop.

Given: u = 28 m s⁻¹, v = 0 m s⁻¹ (it stops), s = 98 m

Acceleration — Eq. (4.4c) links u, v, a and s without needing the time:

Page 41 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

v² = u² + 2as

(0 m s⁻¹)² = (28 m s⁻¹)² + 2 × a × 98 m

0 = 784 m² s⁻² + 196 m × a
a = − 784 m² s⁻² / 196 m

a = − 4 m s⁻ ²

The minus sign says the acceleration is directed opposite to the velocity — the motorbike is
slowing down.
Time taken — now use Eq. (4.4a):

v = u + at

0 m s⁻¹ = 28 m s⁻¹ + (− 4 m s⁻²) × t

4 t = 28 s

t=7s

Check it yourself: s = ut + ½at² = 28 × 7 + ½ × (− 4) × 49 = 196 − 98 = 98 m ✓

Why it happens: choosing Eq. (4.4c) first was the whole trick. The question gives u, v
and s but not t, and (4.4c) is the one kinematic equation in which t does not appear.
Picking the equation that avoids the unknown you do not want saves a round of
substitution every time.

Page 42 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Q6 Fig. 4.27 shows a position-time graph of two objects A and B that are moving along
the parallel tracks in the same direction. Do objects A and B ever have equal
velocity? Justify your answer.

A
B
Position (m)

0 5
Time (s)

Fig. 4.27, page 69 — position–time graphs of two objects A and B moving along parallel
tracks in the same direction. The two lines cross at t = 5 s. (No scale is printed on the
position axis in the book.)

No. Their velocities are never equal.

Page 43 of 75

as e
Class 9 Science Chapter 4 Describing Motion Around Us
a g l AglaSem · NCERT Solutions

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a se
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A B

co m
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m .
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g l a
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.
means the two objects have the same position at that instant — A catches up with B
— not the same velocity. Equal velocity would need equal slopes, which would make
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m as e
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a g l Page 44 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Tip: B starts ahead of A (its line begins higher on the position axis), but A is faster
and overtakes it at t = 5 s. After that A stays ahead and the gap keeps growing.

Q7 A graph in Fig. 4.28 shows the change in position with time for two objects A and B
moving in a straight line from 0 to 10 seconds. Choose the correct option(s). (i) The
average velocity of both over the 10 s time interval is equal since they have the
same initial and final positions. (ii) The average speeds of both over the 10 s time
interval are equal since both cover equal distance in equal time. (iii) The average
speed of A over the 10 s time interval is lower than that of B since it covers a shorter
distance than B in 10 seconds. (iv) The average speed of A over the 10 s time interval
is greater than that of B since B's speed is lower than A's in some segments.
Position (m)

A
B

0 10
Time (s)

Fig. 4.28, page 69 — change of position with time for two objects A and B moving in a
straight line from 0 s to 10 s. Both start together and end at the same position at t = 10
s. (No scale is printed on the position axis in the book.)

The correct options are (i) and (ii).

Page 45 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Position (m)

A

B

0 10
Time (s)
A and B start together and finish together. A keeps a steady velocity; B starts slower and finishes
faster — but the two end points are shared.

Reading the graph: both objects begin at the same position at t = 0 s and arrive at the same
position at t = 10 s. Both move forward the whole time — neither curve ever falls — so neither
turns back.
(i) Correct. Same initial and final positions means the same displacement, over the same 10 s.

average velocity = displacement / time interval

displacement of A = displacement of B ⇒ average velocities are equal

(ii) Correct. Because neither object reverses, the distance travelled equals the magnitude of the
displacement for each of them — and those are equal.

distance travelled = |displacement| (no turning back)

⇒ average speed of A = average speed of B

(iii) Wrong — A does not cover a shorter distance; both cover the same distance in the 10 s.
(iv) Wrong — although B is slower than A in the early part of the interval, it is correspondingly
faster in the later part, and the two effects cancel over the full 10 s.

Page 46 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Why it happens: average speed and average velocity care only about the two end
points and the total time — nothing in between. B's journey is dramatically different
from A's moment by moment (B is accelerating, A is not), but the averages cannot
see that. To distinguish them you would have to compare their velocities at
particular instants, not their averages.

Page 47 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Q8 A truck driver driving at the speed of 54 km h⁻¹ notices a road sign with a speed
limit of 40 km h⁻¹ (Fig. 4.29) for trucks. He slows down to 36 km h⁻¹ in 36 s. What was
the distance travelled by him during this time? Assume the acceleration to be
constant while slowing down.

Page 48 of 75

as e
Class 9 Science Chapter 4 Describing Motion Around Us
a g l AglaSem · NCERT Solutions

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Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Fig. 4.29, page 69 — the speed-limit road sign: 50 km h⁻¹ for cars, 40 km h⁻¹ for trucks.

First convert both speeds to SI units, because the time is in seconds.

u = 54 km h⁻¹ = 54 × 1000 m / 3600 s = 15 m s⁻¹

v = 36 km h⁻¹ = 36 × 1000 m / 3600 s = 10 m s⁻¹

t = 36 s

With constant acceleration the average velocity is the plain mean of u and v, so:

s = ½ (u + v) t

= ½ × (15 m s⁻¹ + 10 m s⁻¹) × 36 s
= ½ × 25 m s⁻¹ × 36 s

= 12.5 m s⁻¹ × 36 s = 450 m

Check it yourself, the long way:
a = (v − u)/t = (10 − 15) m s⁻¹ / 36 s = − 0.139 m s⁻²
s = ut + ½at² = 15 × 36 + ½ × (− 0.139) × 1296 = 540 − 90 = 450 m ✓

Why it happens: the driver is decelerating, so the acceleration is negative — its
direction is opposite to the motion — but the truck is still going forward, so the
distance is positive throughout. Notice that he covers 450 m, most of half a
kilometre, just while shedding 18 km h⁻¹. Speed changes on a highway are not
instant, which is why the speed-limit sign has to be placed well before the stretch it
applies to.

Did you know? He ends at 36 km h⁻¹, which is below the 40 km h⁻¹ limit — so he is
within the law by the end of these 36 s.

Page 50 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Q9 A car starts from rest and accelerates uniformly to 20 m s⁻¹ in 5 seconds. It then
travels at 20 m s⁻¹ for 10 seconds and finally applies the brake (with uniform
acceleration) to stop in 6 seconds. Find the total distance travelled.

Three phases, three areas under one velocity–time graph.

Velocity (m s⁻¹)

20

200 m

50 m 60 m

0 5 15 21
Time (s)
The total distance is the whole area under the graph: a triangle, a rectangle and another triangle.

Phase 1 — speeding up (0 s to 5 s): u = 0, v = 20 m s⁻¹, t = 5 s

s₁ = ½ (u + v) t = ½ × (0 + 20) m s⁻¹ × 5 s = 50 m

Phase 2 — constant velocity (5 s to 15 s): v = 20 m s⁻¹, t = 10 s

s₂ = v × t = 20 m s⁻¹ × 10 s = 200 m

Phase 3 — braking (15 s to 21 s): u = 20 m s⁻¹, v = 0, t = 6 s

s₃ = ½ (u + v) t = ½ × (20 + 0) m s⁻¹ × 6 s = 60 m

Page 51 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

total distance = s₁ + s₂ + s₃

= 50 m + 200 m + 60 m = 310 m

Why it happens: the kinematic equations apply only where the acceleration is
constant, and here it takes three different values (+4 m s⁻², 0, and −10/3 m s⁻²). So
the journey has to be cut at the two points where the acceleration changes, treated
separately, and the results added. The graph makes this visible: three simple shapes,
one after the other.

Check it yourself: braking takes 6 s from 20 m s⁻¹, so a₃ = (0 − 20)/6 = − 3.33 m s⁻².
Then s₃ = 20 × 6 + ½ × (− 3.33) × 36 = 120 − 60 = 60 m ✓

Q10 A bus is travelling at 36 km h⁻¹ when the driver sees an obstacle 30 m ahead. The
driver takes 0.5 seconds to react before pressing the brake. Once the brake is
applied, the velocity of the bus reduces with constant acceleration of 2.5 m s⁻². Will
the bus be able to stop before reaching the obstacle?

Yes — the bus stops 5 m short of the obstacle.

u = 36 km h⁻¹ = 36 × 1000 m / 3600 s = 10 m s⁻¹

Step 1 — the reaction distance. For 0.5 s the driver has not yet touched the brake, so the bus
continues at full speed:

s₁ = u × tr = 10 m s⁻¹ × 0.5 s = 5 m

Step 2 — the braking distance. Now u = 10 m s⁻¹, v = 0, a = − 2.5 m s⁻². Use Eq. (4.4c):

v² = u² + 2as

(0 m s⁻¹)² = (10 m s⁻¹)² + 2 × (− 2.5 m s⁻²) × s₂

0 = 100 m² s⁻² − 5 m s⁻² × s₂
s₂ = 100 / 5 m = 20 m

Page 52 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Step 3 — total stopping distance.

s = s₁ + s₂ = 5 m + 20 m = 25 m

25 m < 30 m ⇒ the bus stops with 5 m to spare

Why it happens: the reaction distance is easy to forget, and it is a fifth of the whole
stopping distance here. During those 0.5 s the brakes do nothing at all — the bus is
simply covering ground at 10 m s⁻¹. Had the driver been distracted for 1.5 s instead
of 0.5 s, the reaction distance alone would be 15 m and the total 35 m: the bus
would have hit the obstacle, with the brakes working exactly as well as before.

Try This: repeat the calculation for a bus at 54 km h⁻¹ (15 m s⁻¹) with the same
reaction time and braking. Reaction distance = 7.5 m, braking distance = 225/5 = 45
m, total 52.5 m — nearly double the obstacle distance. The bus would not stop in
time.

Q11 A student said, “The Earth moves around the Sun”. In this context, discuss whether
an object kept on the Earth can be considered to be at rest.

Both statements can be true at the same time, because rest and motion are always stated with
respect to a chosen reference point.

REFERENCE POINT DOES THE OBJECT'S POSITION CHANGE VERDICT
CHOSEN WITH TIME?

A table, a wall, the ground No The object is at rest

The Sun Yes — it is carried around the Sun with the Earth The object is in
motion

The Earth's axis Yes — it is carried round once a day by rotation The object is in
motion

Page 53 of 75

as e
Class 9 Science Chapter 4 Describing Motion Around Us
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: the chapter's own definition says an object is in motion if its

m l a se
position with respect to the reference point changes with time. Change the reference
o g
point and.cyou can change the answer, without the object doing anything
a
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So a
l serest" is never an absolute property of an object — it is a statement about the
g and the reference point together.
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So the honest answer is: an object kept on the Earth is at rest with respect to the Earth, and in
ag
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motion with respect to the Sun. In everyday emlife we silently take the Earth as our reference
point, which is why we call a book ongal table "at rest" without feeling we are saying anything
incomplete.
a

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Did you know? The Earth carries you around the Sun at about 30 kmm .
s e s⁻¹, and spins

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you eastward
a glaof it, because
com at roughly 0.46 km s⁻¹ at the equator. You feel none
a s em around you — the ground, the air, your chair — moves along with you.
everything

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Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Q12 The velocity-time graph from 0 s to 120 s for a cyclist is shown in Fig. 4.30. Shade
the areas (in different colours) representing the displacement of the cyclist (i)
while cyclist is moving with constant velocity. (ii) when the velocity of cyclist is
decreasing. Also, calculate the displacement and average acceleration in the 120 s
time interval.

6

5
Velocity (m s⁻¹)

4

3

2

1

0 20 40 60 80 100 120
Time (s)

Fig. 4.30, page 70 — velocity–time graph of a cyclist from 0 s to 120 s: 0 → 3 m s⁻¹
between 0 s and 20 s, constant 3 m s⁻¹ from 20 s to 100 s, then down to 2 m s⁻¹ at 120 s.

Reading Fig. 4.30 point by point: the velocity rises from 0 to 3 m s⁻¹ between 0 s and 20 s, stays
at 3 m s⁻¹ from 20 s to 100 s, and falls to 2 m s⁻¹ between 100 s and 120 s.

Page 55 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Velocity (m s⁻¹)
6

5

4

3

2
(i) 240 m
1 (ii) 50 m
30 m

0 20 40 60 80 100 120
Time (s)

Blue: the constant-velocity stretch (20 s – 100 s). Rose: the stretch where the velocity is decreasing
(100 s – 120 s). Green: the initial speeding-up.

(i) Constant velocity — 20 s to 100 s. Shade the rectangle under the flat part of the graph.

displacement = area of rectangle = v × t
= 3 m s⁻¹ × (100 s − 20 s) = 3 × 80 m = 240 m

(ii) Velocity decreasing — 100 s to 120 s. Shade the trapezium under the falling part.

displacement = ½ (sum of parallel sides) × width

= ½ × (3 m s⁻¹ + 2 m s⁻¹) × 20 s = ½ × 5 × 20 m = 50 m

Total displacement in the 120 s. Add the first stretch too (0 s to 20 s, a triangle):

0 s – 20 s: ½ × 20 s × 3 m s⁻¹ = 30 m

20 s – 100 s: 240 m

100 s – 120 s: 50 m

total displacement = 30 m + 240 m + 50 m = 320 m

Page 56 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Average acceleration over the 120 s. This depends only on the velocities at the two ends:

aav = (vfinal − vinitial) / t

= (2 m s⁻¹ − 0 m s⁻¹) / 120 s

= 0.0167 m s⁻² ≈ 0.017 m s⁻², in the direction of motion

Why it happens: the average acceleration is tiny, and positive, even though the
cyclist actually slowed down over the last 20 s. That is because "average" compares
only the first and last readings: she began at rest and ended at 2 m s⁻¹, a net gain.
The graph shows three quite different accelerations along the way (+0.15 m s⁻², 0,
and −0.05 m s⁻²) — the single average hides all of that.

Page 57 of 75

Class 9 Science Chapter 4 Describing Motion Around Us AglaSem · NCERT Solutions

Q13 A girl is preparing for her first marathon by running on a straight road. She uses a
smartwatch to calculate her running speed at different intervals. The graph (Fig.
4.31) depicts her velocity versus time. Estimate the distance she ran based on the
graph.

7.5
Velocity (km h⁻¹)

5.0

2.5

0 2 4 6
Time (h)

Fig. 4.31, page 70 — the runner's velocity–time graph read off the printed grid (one
small square = 0.2 h across and 0.25 km h⁻¹ up).

She runs on a straight road in one direction, so the distance she covers is the area under the
velocity–time graph. Read the seven marked points off Fig. 4.31 (one small square = 0.2 h across
and 0.25 km h⁻¹ up):

TIME (H) 0 0.6 1.6 3.0 4.6 5.6 6.6

VELOCITY (KM H⁻¹) 7.0 7.0 7.5 7.5 7.0 6.5 6.5

Page 58 of 75

as e
Class 9 Science Chapter 4 Describing Motion Around Us
a g l AglaSem · NCERT Solutions

co m
e m.
as
Velocity (km h⁻¹)
m l
m .co a g
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a gl

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a area ≈ 47 km
2.5
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om both axes are in km and h, the area comes out a
. cBecause
m in kilometres.
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Area under the runner's velocity–time graph.

a
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Split the area into six strips and use the trapezium rule for each:

. co
e m
m l as
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0 → 0.6 h: 7.0 × 0.6 = 4.2 km

s m → 1.6 h: ½(7.0 + 7.5) × 1.0 = 7.25 km
e0.6
gl a
a
m
1.6 → 3.0 h: 7.5 × 1.4 = 10.5 km

a se
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3.0 → 4.6 h: ½(7.5 + 7.0) × 1.6 = 11.6 km
m a
ase
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4.6 → 5.6 h: ½(7.0 + 6.5) × 1.0 = 6.75 km

5.6 → 6.6 h: 6.5 × 1.0 = 6.5 km

total ≈ 4.2 + 7.25 + 10.5 + 11.6 + 6.75 + 6.5 = 46.8 km ≈ 47 km
co m
m .
o m l a se
m ag the band 6.5 – 7.5 km
.cCheck it yourself, the quick way: her velocity never leaves
l a se
ag h⁻¹, so it averages about 7 km h⁻¹ over roughly 6.6 h. That gives 7 × 6.6 ≈ 46 km —

.c
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the same answer to the accuracy a graph can give. Since we are reading points by

m a s e
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eye, quoting "about 47 km" is honest; quoting 46.8 km would claim more precision

m .
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than the graph carries.

g l as
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co m
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a g l Page 59 of 75