NCERT Solutions for Class 11 Physics Chapter 2 Motion in a Straight Line

NCERT
SOLUTIONS
CLASS - 11th

aglase .co

Class : 11th
Subject : Physics
Chapter : 3
Chapter Name : Motion In A Straight Line

Q3.1 In which of the following examples of motion, can the body be considered approximately
a point object:
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table

Answer. (a) The size of a carriage is very small as compared to the distance between two
stations. Therefore, the carriage can be treated as a point sized object.
(b) The size of a monkey is very as compared to the size Of circular track. Therefore, the
monkey can be considered as a point sized object on the track.
(c) The size Of a spinning cricket ball is comparable to the distance through which it turns
sharply on hitting the ground. Hence, the cricket ball cannot be considered as a point object.
(d) The size of a beaker is comparable to the height of the table from which it slipped. Hence,
the beaker cannot be considered as a point object.

Page : 55 , Block Name : Exercise

Q3.2 The position-time (x-t) graphs for two children A and B returning from their school O to
their homes P and Q respectively are shown in Figure. Choose the correct entries in the
brackets below :
(a) (A/B) lives closer to the school than (B/A)
(b) (A/B) starts from the school earlier than (B/A)
(c) (A/B) walks faster than (B/A)
(d) A and B reach home at the (same/different) time
(e) (A/B) overtakes (B/A) on the road (once/twice).

Answer. (a) A lives closer to school than B.
(b) A starts from school earlier than B,

(c) B walks faster than A.
(d) A B reach home at the same time.
(e) B overtakes A once on the road.
Explanation: (a) In the given x-t graph, it can be observed that distance OP < OQ Hence, the
distance Of school from the A's home is less than that from B's home.
(b) In the given graph, it can be observed that for x = O, t = O for A, whereas for x = 0, t has
some nite value for B. Thus, A starts his journey from school earlier than B.
(c) In the given x-t graph, it can be observed that the slope of B is greater than that of A. Since
the slope of the x—t gives the speed, a greater slope means that the speed of B is greater than
the speed A.
(d) It Is clear from the given graph that both A and B reach their respective homes at the same
time.
(e) B moves later than A and his/her speed is greater than that of A. From the graph, it is clear
that B overtakes A only once on the road.

Page : 55 , Block Name : Exercise

Q3.3 A woman starts from her home at 9.00 am, walks with a speed of 5 km h on a straight
−1

road up to her of ce 2.5 km away, stays at the of ce up to 5.00 pm, and returns home by an
auto with a speed of 25 km h . Choose suitable scales and plot the x-t graph of her motion.
−1

Answer. Speed of the woman = 5 km/h
Distance between her of ce and home Distance Time taken Speed - 2.5 km
Distance
Time taken =
Speed

2.5
= = 0.5h = 30min
5

It is given that she covers the same distance in the evening bv an auto.
Now, speed of the auto = 25 km/h
Distance
Time taken =
Speed

2.5 1
= = = 0.1h = 6min
25 10

The suitable x-t graph Of the motion Of the woman is shown in the given gure.

Page : 56 , Block Name : Exercise

Q3.4 A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward,
followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and
requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long
the drunkard takes to fall in a pit 13 m away from the start.

Answer. Distance covered with 1 step = 1 m
Time taken = 1 s
Time taken to move rst 5 m forward = 5 s
Time taken to move 3 m backward = 3 s
Net distance covered = 5 - 3 = 2 m
Net time taken to cover 2 m = 8 s
Drunkard covers 2 m in 8 s.
Drunkard covered 4 m in 16 s.
Drunkard covered 5 m in 24 s.
Drunkard covered 8 m in 32 s.
In the next 5 s, the drunkard will cover a distance of 5 m and a total distance of 13 m and a
falls into the pit.
Net time taken by the drunkard to cover 13m = 32 + 5 = 37 s
The x-t graph of the drunkard's motion can be shown as:

Page : 56 , Block Name : Exercise

Q3.5 A jet airplane travelling at the speed of 500 km h ejects its products of combustion at
−1

the speed of 1500 km h relative to the jet plane. What is the speed of the latter with respect
−1

to an observer on the ground ?

Answer. Speed of the jet airplane, v jet = 500 km/h
Relative speed of its products of combustion with respect to the plane, vsmoke = - 1500 km/h
Speed of its products of combustion with respect to the ground = v'smoke
Relative speed of its products of combustion with respect to the airplane,
Vsmoke = v'smoke — vjet
- 1500 = v'smoke - 500
v'srnoke = - 1ooo km/h
The negative sign indicates that the direction Of its products Of combustion is opposite to the
direction of motion of the jet airplane,

Page : 56 , Block Name : Exercise

Q3.6 A car moving along a straight highway with speed of 126 km/h is brought to a stop within
a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does
it take for the car to stop ?

Answer. Initial velocity Of the car, u = 126 km/h = 35 m/s
Final velocity Of the car, v = 0
Distance covered bv the car before corning to rest, s = 200 m
Retardation produced in the car = a
From third equation Of motion, a can be calculated as:
1 2
s11 = ut + at
2

1 2
= 20 × 50 + × 1 × (50) = 2250m
2

Hence, the original distance between the driver of train A and the guard of train B is 2250
-1000 = 1250 m.

Page : 56 , Block Name : Exercise

Q3.7 Two trains A and B of length 400 m each are moving on two parallel tracks with a
uniform speed of 72 km/h in the same direction, with A ahead of B. The driver of B decides to
overtake A and accelerates by 1 m/s . If after 50 s, the guard of B just brushes past the driver
−2

of A, what was the original distance between them ?

Answer. For train A:
Initial velocity, u = 72 km/h = 20 m/s
Time, t = 50s
Acceleration, al =0 (Since it is moving With a uniform velocity)
From second equation of motion, distance (sl)covered by train A be Obtained as:
1 2
s1 = ut + a1 t
2

= 20 × 50 + 0 = 1000m

For train B:
Initial velocity, u = 72 km/h = 20 m/s
Acceleration, a = 1 m/s2
Time,t = 50s
From second equation of motion, distance (sll)cover b}' train A can be obtained as From
second equation of motion, minimum acceleration (a) produced by car B can be obtained as:
1 2
s = ut + at
2

1 2
1000 = 5 × 40 + × a × (40)
2

1600 2
a = = 1m/s
1600

Page : 56 , Block Name : Exercise

Q3.8 On a two-lane road, car A is travelling with a speed of 36 km/h. Two cars B and C
approach car A in opposite directions with a speed of 54 km/h each. At a certain instant, when
the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What
minimum acceleration of car B is required to avoid an accident ?

Answer. Velocity of car A, vA = 36 km/h = 10 m/s
Velocity of car B, vB = 54 km/h = 15 m/s
Velocity of car C, vC = 54 km/h = 15 m/s
Relative velocity of car B with respect to car A,
vBA = vB - vA
= 15 - 10 = 5 m/s
Relative velocity of car C with respect to car A,
vCA = vC - (-vA)
=15 +10 =25 m/s
At a certain instance, both cars B and C are at the same distance from car A i.e.
S = 1 km = 1000 m
Time taken (t) by car C to cover 1000m = 1000/25 = 40s
Hence, to avoid an accident, car B must cover the same distance in a maximum Of 40s.
2 2
v − u = 2as

2 2
(0) − (35) = 2 × a × 200

35×35 2
a = − = −3.06m/s
2×200

From first equation of motion, time (t) taken by the car to stop can be obtained as:

v−u −35
t = = = 11.44s
a −3.06

Page : 56 , Block Name : Exercise

Q3.9 Two towns A and B are connected by a regular bus service with a bus leaving in either
direction every T minutes. A man cycling with a speed of 20 km/h in the direction A to B
notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min
in the opposite direction. What is the period T of the bus service and with what speed
(assumed constant) do the buses ply on the road?

Answer. Let V be the speed Of the bus running between Towns A and B.
Speed of the cyclist, v = 20 km/h
Relative speed of the bus moving In the direction of the cyclist
=V - v = (V - 20) km/h
The bus went past the cyclist every 18 min i.e., h (when he moves in the direction of the
18

60

bus).
Distance covered by the bus (V − 20) 18

60
km

Since one bus leaves after every T minutes, the distance travelled b}' the bus Bill be equal to
T
V ×
60

Both equations (i) and (ii) are equal.

18 VT
(V − 20) × =
60 60

Relative speed of the bus moving in the opposite direction of the cyclist

= (v + 20)km/h

Time taken by the bus to go past the cyclist = 6 min =
6
h
60
6 VT
∴ (V + 20) =
60 60

From equations (iii) and ( iv), we get

6 18
(V + 20) × = (V − 20) ×
60 60

V + 20 = 3V − 60

2V = 80

V = 40km/n

Substituting the value of v in equation (iv), we get

6 40T
(40 + 20) × =
60 60

360
T = = 9 min
40

Page : 56 , Block Name : Exercise

Q3.10 A player throws a ball upwards with an initial speed of 29.4 m/s .
(a) What is the direction of acceleration during the upward motion of the ball ?
(b) What are the velocity and acceleration of the ball at the highest point of its motion ?
(c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point,
vertically downward direction to be the positive direction of x-axis, and give the signs of
position, velocity and acceleration of the ball during its upward, and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player’s
hands ? (Take g = 9.8 m/s2 and neglect air resistance).

Answer. (a) Downward
(b) Velocity = 0, acceleration = 9.8 m/s2
(c) x > O for both up and down motions, v < O for up and v > O for dow > O throughout the
motion
(d) 44.1 m, 6s
Explanation : (a) Irrespective of the direction of the motion of the ball, acceleration (which is
actually acceleration due to gravity) acts in the downward direction towards the centre of the
Earth.
(b) At maximum height, velocity Of the ball becomes zero. Acceleration due to gravity at a
given place is constant and acts on the ball at all points (including the highest point) with
constant value i.e., m/s2.
(c) During upward motion, the sign of position is positive, sign of velocity is negative, and sign
of acceleration is positive. During downward motion, the signs of position, velocity, and
acceleration are all positive.
(d) Initial velocity of the ball. u = 29.4 m/s
Final velocity of the ball, v = O (At maximum height, the velocity of the ball becomes zero)

Acceleration, a = -g = - 9.8 m/s2
From third equation of motion, height (s) can be calculated as:
2 2
v − u = 2gs
2 2
v −u
s =
2g

2 2
(0) −(29.4)
= = 44.1m
2×(−9.8)

From rst equation of motion, time of ascent (t) is given as:
v = u + at

v − u −29.4
t = = = 3s
a −9.8

Time of ascent = Time of descent
Hence, the total time taken by the ball to return to the player’s hands = 3 + 3 = 6s.

Page : 56 , Block Name : Exercise

Q3.11 Read each statement below carefully and state with reasons and examples, if it is true or
false ;
A particle in one-dimensional motion
(a) with zero speed at an instant may have non-zero acceleration at that instant.
(b) with zero speed may have non-zero velocity.
(c) with constant speed must have zero acceleration.
(d) with positive value of acceleration must be speeding up.

Answer. (a) True
(b) False
(c) True
(d) False
Explanation :
(a) VMnen an Object is thrown vertically up in the air, its speed becomes zero at maximum
height. However, it has acceleration equal to the acceleration due to gravity (g) that acts In the
downward direction at that point.
(b) Speed is the magnitude of velocity. When speed is zero, the magnitude of velocity along
with the velocity is zero.
(c) A car moving cn a straight highway with constant speed will have constant velocity, Since
acceleration is de ned as the rate of change of velocity, acceleration of the car is also zero.
(d) This statement is false in the situation When acceleration is positive and velocity is
negative at the instant time taken as origin. Then, far all the time before velocity zero, there is
slowing down of the particle. Such a case happens when pa-ticle is projected upwards, This
statement is true when both velocity and acceleration are positive, at the instant time taken as
origin. Such a case happens when a particle is moving with positive acceleration or falling
vertically downwards from a height

Page : 57 , Block Name : Exercise

Q3.12 A ball is dropped from a height of 90 m on a oor. At each collision with the oor, the
ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.

Answer. Ball is dropped from a height, s = 90 m
Initial velocity of the ball, u = 0
Acceleration, a = g = g.8 rn/s2
Final velocity of the ball = v
From second equation of motion, time (t) taken by the ball to hit the ground can be obtained
as:
1 2
s = ut + at
2

1 2
90 = 0 + × 9.8t
2

t = √18.38 = 4.29s

From first equation of motion, final velocity is given as:

= 0 + 9.8 × 4.29 = 42.04m/s
9 9
Rebound velocity of the ball, ur = v = × 42.04 = 37.84m/s
10 10

Time (t) taken by the ball to reach maximum height is obtained with the help of first

equation of motion as:

′
v = ur + at

′
0 = 37.84 + (−9.8)t

′ −37.84
t = = 3.86s
−9.8

′
Total time taken by the ball = t + t = 4.29 + 3.86 = 8.15s

As the time of ascent is equal to the time of descent, the ball takes 3.86 s to strike back on the
oor for the second time.
The velocity with which the ball rebounds from the oor = 9
× 37.84 = 34.05m/s
10

Total time taken by the ball for second rebound = 8.15 + 3.86 = 12.01 s
The speed-time graph of the ball is represented in the given gure as:

Page : 57 , Block Name : Exercise

Q3.13 Explain clearly, with examples, the distinction between :
(a) magnitude of displacement (sometimes called distance) over an interval of time, and the

total length of path covered by a particle over the same interval;
(b) magnitude of average velocity over an interval of time, and the average speed over the
same interval. [Average speed of a particle over an interval of time is de ned as the total path
length divided by the time interval]. Show in both (a) and (b) that the second quantity is either
greater than or equal to the rst.
When is the equality sign true ? [For simplicity, consider one-dimensional motion only].

Answer. (a) The magnitude of displacement over an interval of time is the shortest distance
(which is a straight line) between the initial and nal positions of the particle. The total path
length of a particle is the actual path length covered by the particle in a given interval of time.
For example, Suppose a particle moves from, point A to point B and then, comes back to a
point, C taking a total time t, as shown below. Then, the magnitude of displacement of the
particle = AC.

Whereas, total peth length AB + BC It is also important to note that the magnitude of
displacement can never be greater than the total path length, However, in some cases, both
quantities are equal to each other.
(b)
Magnitude of average velocity =
Magnitude of displacement

Time interval

For the given particle,
Average velocity = AC \ t
Total path length
Average speed =
Time interval

AB + BC
=
t

Since (AB + SC) > AC, average speed is greater than the magnitude of average velocity The two
quantities Will be equal if the particle continues to move along a straight line.

Page : 57 , Block Name : Exercise

Q3.14 A man walks on a straight road from his home to a market 2.5 km away with a speed of 5
km h–1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5
km h–1. What is the
(a) magnitude of average velocity, and
(b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to
40 min ? [Note: You will appreciate from this exercise why it is better to de ne average speed
as total path length divided by time, and not as magnitude of average velocity. You would not
like to tell the tired man on his return home that his average speed was zero !]

Answer. Time taken by the man to reach the market from home, t 1 =
2.5

5
=
1

2
h = 30min

Time taken by the man to reach home from the market, t 2 =
2.5

7.5
=
1

3
h = 20min

Total time taken in the whole journey = 30 + 20 50 min
Displacement
Average velocity= Time
=
2.5

1
= 5km/h
2

Average speed = = Distance

Time
=
2.5

1
= 5km/h
2

Time = 50 min = 5 / 6 h
Net displacement = 0
Total distance = 2.5 + 2.5 = 5 km
Displacement
Average velocity = Time
= 0

Average speed = = Distance
=
5
= 6km/h …….(b(ii))
Time 5
( )
6

Speed of the man 7.5 km
Distance travelled in rst 30 min = 2.5 km
Distance travelled by the man (from market to home) in the next 10 min
10
7.5 × = 1.25km
60

Net displacement = 2.5 − 1.25 = 1.25km

Total distance travelled = 2.5 + 1.25 = 3.75km
1.25 1.25×3
Average velocity = = = 1.875km/h
40 2
( )
60

3.75
Average speed = = 5.625km/h
40
( )
60

Page : 57 , Block Name : Exercise

Q3.15 In Exercises 3.13 and 3.14, we have carefully distinguished between average speed and
magnitude of average velocity. No such distinction is necessary when we consider
instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to
the magnitude of instantaneous velocity. Why ?

Answer. Instantaneous velocity is given by the rst derivative of distance with respect to time
i.e.,

dx
v ln =
dt

Here, the time interval dt is so small that it is assumed that the particle does not change its
direction of motion. As a result, both the total path length and magnitude of displacement
become equal is this interval of time.
Therefore, instantaneous speed is always equal to instantaneous velocity.

Page : 57 , Block Name : Exercise

Q3.16 Look at the graphs (a) to (d) (Fig. 3.20) carefully and state, with reasons, which of these
cannot possibly represent one-dimensional motion of a particle.

Answer. (a) The given x-t graph, shown in (a), does not represent one-dimensional motion of
the particle. This is because a particle cannot have two positions at the same instant of time.
(b) The given v-t graph, shown in (b), does not represent one-dimensional motion of the

particle. This is because a particle can never have two values of velocity at the same instant of
time.
(c) The given v-t graph, shown in (c), does not represent one-dimensional motion of the
particle. This is because speed being a scalar quantity cannot be negative.
(d) The given v - t graph, shown in (d), does not represent one-dimensional motion of the
particle. This is because the total path length travelled by the particle cannot decrease With
time.

Page : 57 , Block Name : Exercise

Q3.17 Figure shows the x-t plot of one-dimensional motion of a particle. Is it correct to say
from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t
>0 ? If not, suggest a suitable physical context for this graph.

Answer. No
The x-t graph of e particle moving in a straight line for t < O and on a parabolic path for t > O
cannot be shown as the given graph. This is because, the given particle does not follow the
trajectory of path followed by the particle as t = O, x O. A physical situation that resembles the
above graph is of a freely falling body held for sometime at a height.

Page : 58 , Block Name : Exercise

Q3.18 A police van moving on a highway with a speed of 30 km/h res a bullet at a thief’s car
speeding away in the same direction with a speed of 192 km/h. If the muzzle speed of the
bullet is 150 m/s, with what speed does the bullet hit the thief’s car ? (Note: Obtain that speed
which is relevant for damaging the thief’s car).

Answer. Speed of the police Van, vp = 30 km/h = 8.33 m/s
Muzzle speed of the bullet, vb = 150 m/s
Speed of the thief s car, Vt = 192 km/h = 53.33 m/s
Since the bullet is red from a moving Van, its resultant speed can be obtained as:
= 150 + 8.33 = 158.33 m/s
Since both the vehicles are moving in the same direction, the velocity with which the bullet
hits the thief s car can be obtained as:
vbt = vb - vt
= 158.33 - 53.33 = 105 m/s

Page : 58 , Block Name : Exercise

Q3.19 Suggest a suitable physical situation for each of the following graphs ( Figure ):

Answer. (a)The given x-t graph shows that initially a body was at rest. Then, its velocity
increases with time and attains an instantaneous constant value. The velocity then reduces to
zero with an increase in time. Then, its velocity increases with time in the opposite direction
and acquires a constant value. A similar physical situation arises when a football (initially kept
at rest) is kicked and gets rebound from a rigid wall so that its speed gets reduced. Then, it
passes from the player who has kicked it and ultimately gets stopped after sometime.
(b)In the given v-t graph, the sign of velocity changes and its magnitude decreases with a
passage of time. A similar situation arises when a ball is dropped on the hard oor from a
height. It strikes the oor with some velocity and upon rebound, its velocity decreases by a
factor. This continues till the velocity of the ball eventually becomes zero.
(c)The given a–t graph reveals that initially the body is moving with a certain uniform velocity.

Its acceleration increases for a short interval of time, which again drops to zero. This indicates
that the body again starts moving with the same constant velocity. A similar physical situation
arises when a hammer moving with a uniform velocity strikes a nail.

Page : 58 , Block Name : Exercise

Q3.20 Figure gives the x-t plot of a particle executing one-dimensional simple harmonic
motion. (You will learn about this motion in more detail in Chapter 14). Give the signs of
position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s.

Answer. Negative, Negative, Positive (at t = 0.3 s)
Positive, Positive, Negative (at t = 1.2 s)
Negative, Positive, Positive (at t = –1.2 s)
For simple harmonic motion (SHM) of a particle, acceleration (a) is given by the relation:
a = – ω2x ω → angular frequency … (i)
t = 0.3 s
In this time interval, x is negative. Thus, the slope of the x-t plot will also be negative.
Therefore, both position and velocity are negative. However, using equation (i), acceleration of
the particle will be positive.
t = 1.2 s
In this time interval, x is positive. Thus, the slope of the x–t plot will also be positive.
Therefore, both position and velocity are positive. However, using equation (i), acceleration of
the particle comes to be negative.
t = – 1.2 s
In this time interval, x is negative. Thus, the slope of the x–t plot will also be negative. Since
both x and t are negative, the velocity comes to be positive. From equation (i), it can be
inferred that the acceleration of the particle will be positive.

Page : 58 , Block Name : Exercise

Q3.21 Figure gives the x-t plot of a particle in one-dimensional motion. Three different equal
intervals of time are shown. In which interval is the average speed greatest, and in which is it
the least ? Give the sign of average velocity for each interval.

Answer. Interval 3 (Greatest), Interval 2 (Least)
Positive (Intervals 1 & 2), Negative (Interval 3)
The average speed of a particle shown in the x–t graph is obtained from the slope of the graph
in a particular interval of time.
It is clear from the graph that the slope is maximum and minimum restively in intervals 3 and
2 respectively. Therefore, the average speed of the particle is the greatest in interval 3 and is
the least in interval 2. The sign of average velocity is positive in both intervals 1 and 2 as the
slope is positive in these intervals. However, it is negative in interval 3 because the slope is
negative in this interval.

Page : 58 , Block Name : Exercise

Q3.22 Figure gives a speed-time graph of a particle in motion along a constant direction.
Three equal intervals of time are shown. In which interval is the average acceleration greatest
in magnitude ? In which interval is the average speed greatest ? Choosing the positive
direction as the constant direction of motion, give the signs of v and a in the three intervals.
What are the accelerations at the points A, B, C and D ?

Answer. Average acceleration is greatest in interval 2
Average speed is greatest in interval 3

v is positive in intervals 1, 2, and 3
a is positive in intervals 1 and 3 and negative in interval 2
a = 0 at A, B, C, D
Acceleration is given by the slope of the speed-time graph. In the given case, it is given by the
slope of the speed-time graph within the given interval of time.
Since the slope of the given speed-time graph is maximum in interval 2, average acceleration
will be the greatest in this interval.
Height of the curve from the time-axis gives the average speed of the particle. It is clear that
the height is the greatest in interval 3. Hence, average speed of the particle is the greatest in
interval 3.
In interval 1:
The slope of the speed-time graph is positive. Hence, acceleration is positive. Similarly, the
speed of the particle is positive in this interval.
In interval 2:
The slope of the speed-time graph is negative. Hence, acceleration is negative in this interval.
However, speed is positive because it is a scalar quantity.
In interval 3:
The slope of the speed-time graph is zero. Hence, acceleration is zero in this interval.
However, here the particle acquires some uniform speed. It is positive in this interval.
Points A, B, C, and D are all parallel to the time-axis. Hence, the slope is zero at these points.
Therefore, at points A, B, C, and D, acceleration of the particle is zero.

Page : 59 , Block Name : Exercise

Q3.23 A three-wheeler starts from rest, accelerates uniformly with 1 m s–2 on a straight road
for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during
the nth second (n = 1,2,3….) versus n. What do you expect this plot to be during accelerated
motion: a straight line or a parabola?

Answer. Straight line
Distance covered by a body in nth second is given by the relation
a
Da = u + (2n − 1)
2

Where,
u = Initial velocity
a = Acceleration
n = Time = 1, 2, 3, ….. ,n
In the given case,
u = 0 and a = 1 m/s −2

This relation shows that:
Dn ∝ n … (iii)
Now, substituting different values of n in equation (iii), we get the following table:

The plot between n and Dn will be a straight line as shown:

Since the given three-wheeler acquires uniform velocity after 10 s, the line will be parallel to
the time-axis after n = 10 s.

Page : 59 , Block Name : Exercise

Q3.24 A boy standing on a stationary lift (open from above) throws a ball upwards with the
maximum initial speed he can, equal to 49 m/s. How much time does the ball take to return to
his hands? If the lift starts moving up with a uniform speed of 5 m/s and the boy again throws
the ball up with the maximum speed he can, how long does the ball take to return to his hands
?

Answer. Initial velocity of the ball, u = 49 m/s
Acceleration, a = – g = – 9.8 m/s−2

Case I:
When the lift was stationary, the boy throws the ball.
Taking upward motion of the ball,
Final velocity, v of the ball becomes zero at the highest point.
From rst equation of motion, time of ascent (t) is given as:
v = u + at

v−u
t =
a

−49
= = 5s
−9.8

But, the time of ascent is equal to the time of descent.
Hence, the total time taken by the ball to return to the boy’s hand = 5 + 5 = 10 s.
Case II:
The lift was moving up with a uniform velocity of 5 m/s. In this case, the relative velocity of
the ball with respect to the boy remains the same i.e., 49 m/s. Therefore, in this case also, the
ball will return back to the boy’s hand after 10 s.

Page : 59 , Block Name : Exercise

Q3.25 On a long horizontally moving belt (Figure), a child runs to and fro with a speed 9 km/h
(with respect to the belt) between his father and mother located 50 m apart on the moving
belt. The belt moves with a speed of 4 km/h. For an observer on a stationary platform outside,
what is the
(a) speed of the child running in the direction of motion of the belt ?.
(b) speed of the child running opposite to the direction of motion of the belt ?
(c) time taken by the child in (a) and (b) ? Which of the answers alter if motion is viewed by
one of the parents ?

Answer. (a) Speed of the belt, vB = 4 km/h
Speed of the boy, vb = 9 km/h
Since the boy is running in the same direction of the motion of the belt, his speed (as observed
by the stationary observer) can be obtained as:
vbB = vb + vB + 4 = 13 km/h
(b) Since the boy is running in the direction opposite to the direction of the motion of the belt,
his speed (as observed by the stationary observer) carl be obtained as:
vbB = vb + (-vB) = 9 - 4 = 5 km/h
(c) Distance between the child's parents = 50 m
As both parents are standing the moving belt, the speed of the child in either direction as
observed by the parents will remain the same i.e., 9 km/h = 2.5 m/s.
Hence, the time taken by the child to move towards one of his parents is 50

2.5
= 20s

(d) If the motion is viewed by any one of the parents, answers obtained in (a) and (b) get
altered. This is because the child and his parents are standing on the same belt and hence, ere
equally affected by the motion of the belt. Therefore, for both (irrespective of the direction of
motion) the speed of the child remains the same i.e. 9 km h.
For this reason, it can be concluded that the time taken by the child to reach any one of his
parents remains unaltered.

Page : 59 , Block Name : Exercise

Q3.26 Two stones are thrown up simultaneously from the edge of a cliff 200 m high with
initial speeds of 15 m/s and 30 m/s. Verify that the graph shown in Figure correctly represents
the time variation of the relative position of the second stone with respect to the rst. Neglect
air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10
m/s . Give the equations for the linear and curved parts of the plot.
−2

Answer. For rst stone:
Initial velocity, uI = 15 m/s
Acceleration, a = -g = - 10 m/s −2

Using the relation,
1 2
x1 = x0 + u1 ℓ + at
2

Where, height of the cliff, x0 =200 m
2
x1 = 200 + 15t − 5t

when this stone hits the ground, x1 = 0
−5t2 + 15t + 200 = 0

t2 − 3t − 40 = 0

t2 − 8t + 5t − 40 = 0

t(t − 8) + 5(t − 8) = 0

t = 8s or t = −5s

Since the stone was projected at time t = 0, the negative sign before time is meaningless.
T = 8s
For second stone :
Initial velocity, uII = 30 m/s
Acceleration, a = - g = -10 m/s −2

Using the relation,
1
2
x2 = x0 + u11 t + at
2
2
= 200 + 30t − 5t

At the moment when this stone hits the ground: x2=0
−5t2 + 30t + 200 = 0

t2 − 6t − 40 = 0

t2 − 10t + 4t + 40 = 0

t(t − 10) + 4(t − 10) = 0

t(t − 10)(t + 4) = 0

t = 10 s or t = −4s

Hence again, the negative sign is meaningless.

t = 10s

Subtracting equations (i) and (ii), we get

2 2
x2 − x1 = (200 + 30t − 5t ) − (200 + 15t − 5t )

x2 − x1 = 15t

Equation (iii) represents the linear path of both stones. Due to this linear relation between (x2
- xl) and t, the path remains a straight line till 8 s.
Maximum separation between the two stones is at t 8 s. (x2 - x1)rnax 15 x 8 = 120m This is in
accordance with the given graph.
After 8 s, only second stone is in motion whose variation with time is given by the quadratic
equation: x2 -x1 = 200 + 30t - 5t^2 Hence, the equation of linear and curved path is given by
x2 - x1 = 15t (Linear path)
x — x1 = 200 + 30t — 5t (Curved path)
2 2

Page : 59 , Block Name : Exercise

Q3.27 The speed-time graph of a particle moving along a xed direction is shown in Figure.
Obtain the distance traversed by the particle between (a) t = 0 s to 10 s, (b) t = 2 s to 6 s.

What is the average speed of the particle over the intervals in (a) and (b) ?

Answer. (a) Distance travelled by the particle = Area under the given graph
1
= × (10 − 0) × (12 − 0) = 60m
2

Distance 60
= = 6m/s
Time 10

(b) Let s1 and s2 be the distances covered by the particle between time
T = 2s to 5s and t = 5 s to 6 s respectively.
Total distance (s) covered by the particle in time t = 2 s to 6 s

For distance s1:
Let u' be the velocity of the particle after 2 s and a' be the acceleration of the particle in t = 0 to
t = 5 s.
Since the particle undergoes uniform acceleration in the interval t = 0 to t = 5 s, from rst
equation of motion, acceleration can be obtained as:
v = u + at
where,
v = Final velocity of the particle
12=0 + a' x 5
′ 12 2
a = = 2.4m/s
5

Again, from rst equation of motion, we have

V = u + at
= 0 + 2.4 x 2 = 4.8 m/s
Distance travelled by the particle between time 2 s and 5 s i.e., in 3 s
1
′ ′ 2
s1 = u t + a t
2
1
2
= 4.8 × 3 + × 2.4 × (3)
2

= 25.2m

For distance s2:
Let a” be the acceleration of the particle between time t = 5 s and t = 10 s.
From rst equation of motion,
v = u + at ( where v = 0 as the particle finally comes to rest)
′′
0 = 12 + a × 5

−12
∗
a =
5
2
= −2.4m/s

Distance travelled by the particle in 1 s (i.e., between t = 5 s and t = 6 s)
1
′′ 2
s2 = u t + at
2
1
2
= 12 × a + (−2.4) × (1)
2

= 12 − 1.2 = 10.8m

From equations (i), (ii), and (iii), we get

s = 25.2 + 10.8 = 36m

36
∴ Average speed = = 9m/s
4

Page : 60 , Block Name : Exercise

Q3.28 The velocity-time graph of a particle in one-dimensional motion is shown in Figure

Which of the following formulae are correct for describing the motion of the particle over the
time-interval t2 to t1 ?
(a) x(t2 ) = x(t1 ) + v (t1 ) (t2 – t1 ) +(½) a (t2 – t1 )
2

(b) v(t2 ) = v(t1 ) + a (t2 – t1 )
(c) vaverage = (x(t2 ) – x(t1 ))/(t2 – t1 )
(d) aaverage = (v(t2 ) – v(t1 ))/(t2 – t1 )
(e) x(t2 ) = x(t1 ) + vaverage (t2 – t1 ) + (½) aaverage (t2 – t1 )
2

(f) x(t2 ) – x(t1 ) = area under the v-t curve bounded by the t-axis and the dotted line shown.

Answer. The correct formula describing the motion of the particle are (c), (d) and, (f) The
given graph has a non-uniform slope. Hence, the formulae given in (a), (b), and (e) cannot
describe the motion of the particle. Only relations given in (c), (d), and (f) are correct
equations of motion.

Page : 60 , Block Name : Exercise