JEE NEET Physics Question Bank - Kinematics

2 Kinematics
Vector
Equal Parallel Opposite Antiparallel Aparallel
vectors vectors vectors vectors vectors
Magnitude Equal Unequal Equal Unequal Equal or Unequal
Direction Equal Equal Opposite Opposite any
Angle between q = 0° q = 0° q = 180° or q = 180° or q ¹ 0°, q ¹ 180°
two vectors p rad p rad 0 < q < 360°
ˆ When a vector is multiply by any value or scaler, then direction remain as it is but magnitude
becomes scaler times.
Addition or Subtraction of vectors
ˆ Method of triangle
ˆ Method of parallelogram and
ˆ Algebraic or analytical Method
It is resultant vector of addition of vector and vector B , then
  
ˆ R A
o
|R | = A 2  B2  2ABcos T .

and R makes an angle a with vector A , then
 

Bsin T R
tan a =
A  Bcos T
= y  
A+
Rx B  B
 =
ˆ Magnitude of unit vector is one unit, it is expressed as n . R

Unit vector in direction of A ,

ˆ q a
o 
n A = Vector A
=
A
o magnitude of vector
|A |
ˆ Unit vectors in three dimensions (i.e. cartesian co-ordinate system) X-axis ® i , Y-axis ® j ,
Z-axis ® k Y R
Y
Resolution of a vector Ax
P P
Az 
A
Ay Ay Ay
O

A Y
Ax Az M
Ax
X
O Ax Ay Q
X
In two dimension In three dimension

A = Ax i + Ay j and A = Ax i + Ay j + Az k
 

o o
|A| A x 2  A y2 |A| Ax 2  A y2  Az 2

18

ˆ Angular resolution of a vector (In two dimensions) Y
A = A cosq i + A sinq j


ˆ
ˆ A vectors makes an angle q with x-axis in two dimensions, then Q
S
§ Ay ·

A sinq
\ q = tan–1 ¨ ¸
Ay A
tanq =
Ax © Ax ¹ R q
A cosq
ˆ Subtraction of vectors : P
To make subtraction, add an opposite vector of which will
be subtract, in given vector. O X
N A cosq M


A

– 
 B
A =

R = A + (– B)
  
–B

R = A – B
  

If R = A – B then
  
ˆ
o  Bsin T
|R | = A 2  B2 – 2ABcos T and an angle of R with A is a then, tan a = A  Bcos T
 

Multiplication of vectors

Scaler product Vector Product
(dot product) (cross product)

o o o o
ˆ A˜ B = | A | | B | cos q ˆ Au B = | A | | B | sin q n
   

o o
ˆ A˜ B = The product of the magnitude of first Where n is unit vector in the direction
vector with the projection of second vector on obtained by right handed screw rule.
the first vector
o o oo o o o o
ˆ A˜ B = B˜ A ˆ A u B = – Bu A
§ o o· § o o·
u ¨ Au C ¸
o o o
ˆ A u ( B  C) = ¨ ¸ +
o o o o o o o
© ¹
ˆ A ˜ ( B  C) A ˜ B  A ˜ C
© ¹
A B

ˆ If A and B are parallel then, ˆ If A and B are parallel or anti parallel
   

o o § o o · , then o o
A˜ B = | A | | B | ¨A &B¸ Au B = 0
 

©  ¹ 
§ o o·
ˆ If A and B are antiparallel, then If A and B are perpendicular ¨ A A B ¸
 
ˆ
© ¹
o o
A˜ B = – | A | | B | then, o o = | A | | B | n
   
Au B

19

i j k
o o
ˆ If A and B are perpendicular ˆ Au B =
  Ax Ay Az
Bx By Bz
o o
then A ˜ B = 0
o o
ˆ A˜ B = AxBx + AyBy + AzBz ˆ Area of the triangle formed by A and B ,
 

1  
D= |A×B|
2
ª º
A x Bx  A y B y  A z Bz Sin D Sin E Sin J
–1 « »
ˆ q = cos « 2 »
= =
 
ˆ C a
«¬     »¼
2 2 2 2 2 B
( A x A y A z ) ( B x B y B z ) A B C
b g

A
(1) Magnitude of A and B are 10 unit and 20 unit respectively. If the angle between this two vectors is
 

30°, the magnitude of resultant (R) is ...... unit and angle with vector A is ...... ( 3 = 1.7).


(A) 30 unit, 30° (B) 10 unit, 30° (C) 29 unit, 20° 19' (D) 39 unit, 30° 30'

(2) A force act on a substance having mass m is F = 5 i – 4 j + 2 k N. An angle of this force


with X-axis is ......

§ 1 · § 5· § 3 ·
(A) Cos–1 ¨ ¸ (B) Cos–1 ¨¨ 3 ¸¸ (C) Cos–1 ¨ ¸ (D) Cos–1 3 5
©3 5 ¹ © ¹ © 5¹

o o
(3) | A × B | is greater by 73 % than A ˜ B for given A and B . An angle between
   

A and B is ......
 

(A) 0° (B) 30° (C) 60° (D) 90°

(4) If the addition vector R of A and B makes an angle a with A and b with B . Then ......
    

(A) Always a < b (B) If A < B, a < b
(C) If A = B, a < b (D) If A > B, a < b
 
(5) An angle between A = 3 i – 4 j – 2 k and B = 8 i + 3 j + 6 k is ......
(A) 90° (B) 60° (C) 180° (D) 0°
 
(6) If the vectors A = 5 i + 7 j – 3 k and B = – 2 i + m j + 6 k are mutually perpendicular,
then m = ......
(A) 1 (B) 4 (C) 3 (D) –4
  JG JG
(7) If A = 2 i – j + 3 k and B = i + 2 j – k then unit vector in the direction of A u B is ......
1
(A) (3 i + j +2 k ) (B) i – j + k
14
1 1
(C) ( i + j + k ) (D) (– i + j + k )
3 3
20


(8) Area of triangle formed by A = 4 i + 3 j – k and B = i + 2 j + k is ...... unit.


(A) 10 3 (B) (C) (D) 5 3
5 3 3
2 2

(9) | A × B |2 + | A × B |2 = ......
   

(A) zero (B) AB (C) AB (D) A2B2

(10) If | A + B | = | A | = | B |, then the angle between A and B is ...... .
     

(A) 0° (B) 30° (C) 90° (D) 120°

(11) The unit vector, which is perpendicular to A and B both is ...... .
 

o o o o o o o o
AB Au B Au B A˜ B
(A) (B) (C) (D)
AB ABsin T ABcos T ABsin T
(12) If 23 i + 13 j + m k is unit vector, then m = ...... .


(A) 3 (B) – 13 (C) 1 (D) 13
2

o o
(13) If | A | = 4, | B | = 2.5, A ˜ B = 8 and an angle between A and B is acute angle then
   

o o
| A u B | = ....... .
(A) 6 (B) 10 (C) 3.2 (D) 2

(14) Aproximate area of parallelogram fromed by A = 2 i – 3 j – k and B = 4 i + 3 k is ...... unit.


(A) 5 (B) 11 (C) 18 (D) 15

(15) The component of A = 3 i – 2 j + k in the direction of B = i + j is ...... .


(A) 1 (B) (C) (D)
1
2 2 14
Ans : 1 (C), 2 (B), 3 (C), 4 (D), 5 (A), 6 (B), 7 (D), 8 (B), 9 (D), 10 (D), 11 (B), 12 (A),
13 (A), 14 (C), 15 (C)

Mechanics
ˆ A branch of physics dealing with properties of a moving body, causes of motion, results due to
motion etc. is known as mechanics.
ˆ Mechanics devides in two branches.
ˆ Kinematics : A branch of mechanics dealing with motion without considering its causes is
known as kinematics.
ˆ Dynamics : A branch of mechanics describing motion along with its causes and properties
of moving body is called dynamics.
Concept of a particle and frame of reference :
ˆ A point like object having mass can be considered as a particle.
ˆ If the distance between two object is very large as compared to their dimensions, these objects
can be treated as particle.

21

ˆ All the particles of a solid body performing linear motion cover the same distance in the same
time. Hence, the motion of such a body can be described in terms of the motion of any of its
constituent particles.
ˆ Particle is relative concept.
ˆ Motion is also a relative concept.
ˆ Motion is described according to a place and a situation of an observer.
ˆ A place and a situation from where an observer takes his observation is called frame of reference.
Frame of reference is arbitrary.
ˆ Frame of reference will be inertial or non-inertial.
Position, Path length and Displacement :
ˆ To describe a motion and position of particle, we can choose three mutually perpeducular axis
name them X, Y and Z.
ˆ If all the three co-ordinates of a particle remain unchanged with time, the particle is considered
at rest with respect to time in this frame. If one or more co-ordinates of a particle changes
with time, we say that particle is in motion with respect to this frame.
Path length Displacement
ˆ The total distance travelled by a particle in ˆ The change in position of a particle in some
some time interval is called the path length time interval is called displacement.
or total distance.
ˆ Path length depends on path. ˆ Displacement does not depends on path. Its
depends only on initial and final position.
ˆ Path length is always positive. ˆ Displacement may be positive, negative or zero.
ˆ Path length is scaler quantity. ˆ Displacement is vector quantity.
ˆ In a normal condition, ratio of path length to displacement is one or more than one.
(16) In a joggers park, a man complete two rotation on a circular path of radius 30 m. Then
difference of path length and displacement is ...... m in given motion.
(A) 120 p (B) 60 p (C) zero (D) 30 p
(17) A person travelled a distance of 40 m along a straight line, then he travelled 30 m in right side.
So, the ratio of total distance to displacement is ......
(A) 1 (B) 1.33 (C) 0.75 (D) 1.4
(18) An ant experienced 60 cm displacement on a cicular path of 60 cm radius. Then total distance
travelled by it is ...... cm.

S
(A) (B) 20 p (C) 60 p (D) 120 p
3

Ans. : 16 (A), 17 (D), 18 (B)

Average speed and average velocity :

ˆ With help of average speed, we know how fast the position of object is changing.

ˆ The ratio of total distance travelled (i.e. path length) to the time interval during which the
motion has been taken place is known as average speed. Thus, distance travelled by object in
unit time is known as average speed.

22

Total travelled distance Path length
average speed = <v> = v = =
time interval time interval
ˆ Average velocity is the ratio of displacement to the time interval, in which the displacement
occurs.Thus, the displacement covered by object in unit time is known as average velocity.

displacement
average velocity =
time interval

ˆ Unit of average speed and average velocity is ms–1.
ˆ Average speed is scaler while average velocity is vector physical quantity.

(19) A vehicle travels different distances with different speeds in the same direction. The average
speed of the vehicle is ......
v1  v2  v3  .....
(A) v1+ v2+ v3 + .... (B) t  t  t  ....
1 2 3

d1 d2 d3 d1  d 2  d3  ....
(C) t + t + t + ..... (D)
1 2 3 d1 d 2 d3
   ....
v1 v2 v3

(20) A cyclist travelled third part of total distance with a speed of 12 kmh–1 and remaining distance
with a speed of 18 kmh–1. so, average speed of his is ...... kmh–1.
(A) 15 (B) 12 (C) 18 (D) 15.43
(21) A person wallking in a straight line, covers 13 part of the distance to be travelled with a speed of
v1 and remaining distance with speed v2. So, avrage speed of his is ......
3v1 v2 3v1 v2 2v1 v2 2v1 v2
(A) v  2v (B) 2v  v (C) v  2v (D) 2v  v
1 2 1 2 1 2 1 2
(22) A car covered 120 km straight line disance in 2 hour, then it terned left and travelled 50 km
distance in 1 hour. So average speed and average velocity during this is ...... and ......
respectively.
(A) 40 kmh–1, 50 kmh–1 (B) 56.67 kmh–1, 43.33 kmh–1
(C) 43.33 kmh–1, 56.67 kmh–1 (D) 50 kmh–1, 40 kmh–1
(23) A length of second arm of a clock is 10 cm. This second arm is moves to 9 from 3. Average
speed and average velocity of this second arm is ...... cms–1 and ...... cms–1 respectively.
(A) 3.33, 3.33 (B) 1.05, 0.6666 (C) 1.111, 3.00 (D) 1.666, 3.33

Ans. : 19 (D), 20 (D), 21 (B), 22 (B), 23 (B)
Instantaneous velocity and Instantaneous speed :
ˆ Instantaneous velocity is obtained by time derivative of position.
ˆ Instantaneous velocity is vector physics quantity and its direction is in direction of tangent
of path.

23

ˆ Magnitude of instantaneous velocity is instantaneous speed.
ˆ In a graph of x ® t, instantaneous speed is slope of tangent at given point of path.

(24) A motion of object is expressed by x(t) = 2t2 – 3t + 4 m. Average speed of object for initial 3 s
and instantaneous speed at t = 3 s is ...... and ...... respectively.

(A) 3 ms–1, 9 ms–1 (B) 9 ms–1, 9 ms–1

(C) 4.5 ms–1, 3 ms–1 (D) 4.5 ms–1, 4.5 ms–1
(25) An equation of motion for a particle is given by v = 2t + 3 ms–1. So, distance travelled by particle
in initial four second is ......
(A) 11 m (B) 3 m (C) 19 m (D) 28 m

(26) An object is moving on a circular path with velocity v at a given instant. When it completes half
rotation change in its velocity will be ......

(A) zero (B) – v (C) –2 v (D) 2 v
 

(27) Motion of an object in a straight line is given by x = 5t2 – 6t + 2.2 m. The position of the object
when its velocity is zero is given by ...... m.
(A) 0.6 (B) 0 (C) 0.4 (D) 1.0
Ans. : 24 (A), 25 (D), 26 (C), 27 (C)
ˆ Acceleration :
ˆ The time rate of change of velocity of moving body is called acceleration.
change in velocity
Average acceleration =
time interval
a =<a>=
' v
't
o
o
Instantaneous acceleration a = d v = time derivative of velocity = v

dt

o o
d2 x
= = Second derivative of position with respect to time = x

dt 2
(28) A motion of particle is given by x(t) = 4t2 + 4t –3 m. So, instantaneous acceleration of particle
for 4th second will be ...... ms–2.
(A) zero (B) 8 (C) 4 (D) 16
(29) For a moving particle, the relation between time and position is given by t = 4x2 + 3x. Where t is in
second and x is in meter. So, acceleration for this particle a = ...... . (as a function of velocity).
(A) 12v (B) 8v2 (C) –8v3 (D) –12v2
(30) For moving particle, If x = At2 + Bt + C then equation for velocity ...... and quation for
acceleration ...... .
(A) 2At + B, 2A (B) 2t + B, 2B (C) At + B, 2A (D) At2, Bt

24

(31) The velocity of a particle becomes 9 i + 2 j ms–1 from (4 i – 3 j ) ms–1 in 5 second. Then
average acceleration is ...... ms–2.
(A) i + j (B) 13 i – 5 j (C) 5 i – 5 j (D) i – j
(32) An acceleration of moving particle is a = 4t ms–2. So, velocity at third second will be ...... ms–1.
(A) 12 (B) 18 (C) 36 (D) zero
Ans. : 28 (B), 29 (C), 30 (A), 31 (A), 32 (B)
Kinetic equations for uniformly acceleration motion :
At t = 0 position of particle x0 = 0 then,
at t = t velocity of particle is v = v0 + at.

displacement of particle is x = v0t + 12 at2

§ v  v0 ·
x = ¨ 2 ¸t
© ¹

v 2  v02
x=
2a
In general, at t = 0, position of particle x = x0 then
v = v0 + at

x – x0 = v0t + 12 at2

§ v  v0 ·
x – x0 = ¨ 2 ¸ t
© ¹

v 2  v02
x – x0 =
2a
A distance travelled in nth second by a perticle, which moves with constant acceleration and v0 as
initial velocity.

dnth = v0 + 2 (2n –1)
a

(33) A car which stands at traffic signal, starts its motion with acceleration 5 ms–2. Then velocity of
car after 4 s and distance travelled by car during this 4 s will be ...... and ...... respectively.
(A) 40 ms–1, 20 m (B) 20 ms–1, 20 m (C) 40 ms–1, 40 m (D) 20 ms–1, 40 m
(34) The ratio of the distances travelled in the fifth and the third second by a particle moving start
from rest over a straight path with constant acceleration 8 ms–2 is ......

(A) 53 (B) 95 (C) 10
3
(D) 95
(35) An object starts its constant accelerated motion with 1 ms–1 initial velocity and it covered same
distance in third second as it covered in initial two seconds. Then constant acceleration of object
during this motion is ...... ms–2.
(A) 1 (B) 5 (C) 2 (D) 3
(36) The velocity of an object moving with constant acceleration is 45 ms–1 at the end of 10 s and
becomes 53 ms–1 at the end of 12 second. Then the distance travelles by object in 15 second will
be ...... m.
(A) 65 (B) 82.5 (C) 110 (D) 525

25

(37) The object srarts from stationary, moves with constant acceleration 4 ms–2 for 10 s then it moves with
constant velocity for 50 s and finally it stops with 2 ms–2 deceleration. Total distance coverd by object
will be ......
(A) 1200 m (B) 1300 m (C) 2000 m (D) 2600 m

(38) In a constant acceleration motion, an object losing 23 v0 velocity in time t. Where v0 is initial

velocity. So time required to velocity becomes zero will be ...... .

(A) t (B) 2t (C) 2 (D) 3
3t 2t

(39) Speeds of on object moving with constant acceleration at point A and B are u and v
respectively. Then the speed of object at midpoint of A and B is ...... .

v2  u 2 vu v2  u 2 v2  u 2
(A) (B) (C) (D)
2 2 2 2
(40) An object starts its motion with 2 ms–1 velocity and acceleration given by a = 6t2 – 2t + 3 ms–2.
So, velocity of object at t = 3 s is ...... ms–1.
(A) 20 (B) 56 (C) 48 (D) 51
Ans. : 33 (D), 34 (B), 35 (C), 36 (D), 37 (D), 38 (C), 39 (A), 40 (B)

Kinetic equations for freely falling body :

ˆ For freely falling body v0 = 0 and a = –g.
upward direction ® Positive y-axis ® Positive.

downward direction ® negative y-axis ® negative.

ˆ For freely falling body,
v = – gt –2gy = v2

y = – 12 gt2 – y = 12 vt

(41) Two different substances are allowed to free falling from towers of height h1 and h2. Time taken

by them to reach on ground are t1 and t2 respetively. So t = ......
t1
2

h12
(A) h (B) (C) (D)
h1 h1 h2
2 h2 h2 2 h1

(42) An object is allowed to fall freely from a tower of 80 m height. Distance covered by it in last
second before touching ground is ...... m. (g = 10 ms–2)

(A) 4 (B) 45 (C) 35 (D) 10

(43) A freely falling body passed a building of height 25 m in last second. Then from ...... m height
that body will be allowed to free fall. (g = 10 ms–2)

(A) 125 (B) 100 (C) 85 (D) 45

26

(44) The time taken to reach on ground by a freely falling object from height h1 and h2 are t1 and t2

respectively then t = ...... .
t2
1

h 22
(A) h (B) (C) (D)
h2 h2 h1
1 h1 h12 h2

(45) An object is thrown from a tower of height h in vertically upward direction with speed v. So time
taken by object to touch a ground is ......

v ª 2 hg º
(A) 1+ (B) g 1+ (C) g 1– (D) g «1+ 1+ »
2g h v hg v 2 hg
v2 v2 v2 «¬ v 2 ¼»
(46) A stone is dropped from a balloon, starts its motion with 4 ms–2 in vertically upward direction
after 5 s. So, minimum height from ground obtain by this stone is ...... m. (g = 10 ms–2)
(A) 55 (B) 70 (C) 90 (D) 105
(47) An object is thrown from a tower of height H in vertically upward direction with velocity v0,
when it touch the ground its velocity becomes 2v0. then H = ...... .
3 v02 3 v02 v 02
(A) 2 g (B) (C) 2 g (D) 2 g
v0
g
(48) An object is allowed to fall freely from a tower of hight 150 m, exactly at the same time another
object is thrown from the bottom of the tower in vertically upward direction with velocity of
30 ms–1. So, two object will be meet after ...... time.
(A) 10 s (B) 45 s (C) 5 s (D) 12 s
(49) A ball is allowed to fall freely from 10 m height from a surface. The ball reaches at 5 m height
after colliding with surface. If time of contact for ball and surface is 0.05 s then average
acceleration during contact is ...... . (g = 10 ms–2)
(A) 241.4 ms–2 (B) 965.2 ms–2 (C) 4828 ms–2 (D) 482.8 ms–2
Ans. : 41 (B), 42 (C), 43 (D), 44 (B), 45 (D), 46 (B), 47 (A), 48 (C), 49 (D)
Relative velocity :
ˆ velocity of P with respect to A =vPA, velocity of P with respect to B = vPB, velocity of B with
respect to A = vBA.
According to this vPA = vPB + vBA (B will be eliminated.)

ˆ Thus, velocity of P with respect to A is obtained by addition of velocity of P with respect to B and
velocity of B with respect to A.
ˆ Above matter can be expressed as below :
vBA = vPA – vPB
= –vAP – (–vBP)
= vBP – vAP (P will be eliminated) vBA = vB – vA

Thus relative velocity vBA of moving body B with respect to moving body A will be obtained by
subtraction of vBP (velocity of moving body B with respect to any stationary object or point P) and
vAP (velocity of moving body A with respect to any stationary object or point P.)
Stationary object means reference object or point will be taken any object or point.

27

(50) A truck passes through a traffic signal with a constant speed 15 ms–1 while at same instant, a car
start its motion with constant acceleration 4 ms–2 in the direction of truck’s motion. So, relative
velocity of car with respect to truck after 6 sec is ...... .
(A) 9 ms–1 (B) –9 ms–1 (C) 39 ms–1 (D) –39 ms–1
(51) Car A is moving on a road with constant speed 54 kmh–1. Now car B and car C are moving with
72 kmh–1 in mutually opposite direction on the same road. At any cretain time (instant), distance
of car B and car C from car A is 1.4 km, driver of car B think to overtake car A. So,
acceleration of car B will be ...... .
(A) 1 ms–2
Car B Car A Car C
(B) 1.5 ms–2
(C) 2 ms–2
(D) –1 ms–2

(52) The rain is falling verticully downward with speed of 12 kmh–1. A man running with 5 kmh–1 on a
straight way. velocity of rain experience by man will be ...... .
(A) 7 kmh–1 (B) 13 kmh–1 (C) 17 kmh–1 (D) 119 kmh
–1

(53) The speed of boat is 5 kmh–1 in still water. It crosses a river of width 1.0 km along the shortest
possible path in 20 min. Then, speed of river will be ...... kmh–1.
(A) 1 (B) 3 (C) 4 (D) 5
(54) A stationary man observes that the rain is falling vertically downward. When he starts running
with a velocity of 10 kmh–1, he observes that the rain is falling at an angle 30° with the vertical
direction. The actual velocity of rain is ...... .

(A) 10 3 kmh–1 (B) kmh–1 (C) 5 3 kmh–1 (D) 15 kmh–1
10
3

Ans : 50 (A), 51 (B), 52 (B), 53 (C), 54 (A)

Relative displacement :
At t = 0, relative positions of particle A and B with respect to reference point (origin) are xA0 and
xB0 and relative velocities of particles A and B with respect to reference point are vA and vB. So,
relative displacement of particle B with respect to particle A at t = t will be,
xB – xA = (xB0 – xA0) + (vB– vA) t
xBA vBA
¿
°
¾
°
½

¿
°
¾
°
½

­ ­

Position of particle B with relative velocity of particle B
respect to particle A at t = 0 with respect to particle A.

(55) At t = 0, car A and car B are at 50 m and 130 m away from origin respectively. Both car starts
their motion in same direction with constant velocities 15 ms–1 and 7 ms–1 respectively. When and
where they overtake each other ?
(A) 10 s, 200 m (B) 15 s, 170 m (C) 8 s, 180 m (D) 12 s, 150 m

28

(56) When particles A and B move uniformly towards each other, they get 16 m closer in 2 second
and when they move uniformly in same direction with original speeds they get 16 m closer in 8
second. So, the speeds of these particles will be ...... and ...... .
(A) 5 ms–1 and 3 ms–1 (B) 8 ms–1 and 0.5 ms–1
(C) 4 ms–1 and 4 ms–1 (D) 32 ms–1 and 2 ms–1
Ans. : 55 (A), 56 (A)
Motion in one, two and three dimensions :
Motion in one motion in two motion in three
dimension dimension dimension
Position x r = xi + y j r = xi + y j + zk
    
 

o o o o
displacement Dx = x – x0 ' r = r – r0 ' r = r – r0
 

= (x – x ) i + (y – y ) j
0 0
= (x – x ) i + (y – y ) j
0 0

+(z – z0) k
o
o o
o
'x
average speed <v> = ' t | v | = |' r | | v | = |' r |
't 't
o o
Instantaneous v = dt v = dt v = dt
dx  dr  dr

velocity
'v o
o o
o
average <a> = ' t 'v 'v
a! = 't a! = 't
acceleration
2 o o d2 
Instantaneous a = dt = 2 = dv = d 2r a = dt ( v ) = ( )
dv d x  2  d 
dt a dt dt 2 r
acceleration
dt

 
Kinematic v = v0 + at v = v0 + a t v = v0 + a t
   

equations of x = v0t + 12 at2 d = v t+ 2 at d = v t + 12 a t2
 
2
 
1 

§o o·
0 0
§o o·
§ v  v0 · ¨ v  v0 ¸ ¨ v  v0 ¸
uniformly accel- x = ¨ 2 ¸t d = ¨ 2 ¸t = ¨ 2 ¸t
© ¹
 
¨ ¸ ¨ ¸
d
© ¹ © ¹
ereted motion

2ax = v2 – v02 2 a . d = v2 – v02 2 a . d = v2 – v02
   

ˆ Dircetion of instantaneous velocity is in a direction of tangent drawn at given point of path.

Instantaneous velocity in two dimensions is v = vx i + vy j , then an angle made by instantaneous

ˆ
vy
velocity with x-axis is obtained by equation. tan q = v
x
ˆ Velocity is vector physical quantity, hence it can change in three ways :
(1) Magnitude of velocity will change. (In a direction of velocity)
(2) Direction of velocity will change. (In a perpendicular direction to velocity)
(3) Magnitude and direction both will change. (In direction and perpendicular direction of velocity)
ˆ Acceleration is result of change in velocity, hence it results three different ways.
ˆ If only magnitude of velocity changes then acceleration is parallel to velocity. It known as a||.

29

ˆ If only direction of velocity changes then acceleration is perpendicular to velocity. It known as a^.
ˆ If magnitude and direction of velocity both changes then acceleration will be parallel and
perpendicular to velocity. so,
o o o o
a = a & + aA Here a and a A are mutually perpendicular to each other.

&

o
\|a|= a&2  aA 2

ˆ During uniform circular motion, only direction of velocity changes. The direction of acceleration is
perpendicular to the direction of velocity, it means in the direction of centre of circular path. That’s
why this acceleration is called Radial acceleration or Centripetal acceleration (ar or ac.)

ar = ac = v
2
r

(57) The position vector of a moving particle depends on time according to r = at2 i + (bt – 3) j m,


then magnitude of velocity and magnitude of acceleration at time t will be ...... and ......
respectively.
(A) at2 + bt – 3, at2 (B) 2at + b, 2a (C) 4D 2 t 2 + E2 , 2a (D) 2at + b – 3, 2at
(58) Ships A and B are standing 20 km far from each other on a line joining north south. Ship A
moves with 10 kmh–1 in west direction and ship B moves with 10 kmh–1 in north direction. Then
minimum distance between these two ships will be ...... km obtain after ...... minute.

(A) 10 2 , 60 (B) 20 2 , 60 (C) , 15 (D) , 15
10 20
2 2
(59) A particle starts its motion from the origin with velocity 3 j ms–1 and moves in xy plane with


uniform acceleration 2 i + j ms–2. When the value of x co-ordinate is 25 m, then the value of
y co-ordinate will be ...... m.
(A) 25 (B) 54 (C) 27.5 (D) 55
Ans. : 57 (C), 58 (A), 59 (C)

Projectile motion
When an object is thrown in gravitational field of earth it moves with constant horizontal velocity and
only gravitational acceleration. Such two dimensional motion is called a Projectile motion.
For projectile motion,
Horizontal direction Vertical direction
v0x = v0cos q0 = constant v0y = v0 sinq0
ax = 0 ay = –g
Position at time t, x = (v0cosq0)t Position at time t, y = (v0sin q0)t – 1 gt2
2

velocity at time t, vx = v0 cosq0 velocity at time t, vy = v0 sin q0 – gt

Position of projectile at time t, r = (v0 cosq0)t i + [(v0 sinq0)t – 1 gt2] j

ˆ 2

velocity of projectile at time t, v = v0 cosq0 i + (v0 sinq0 – gt) j


30

Equation of trajectory of a Projectile, y = (tanq0)x – x2
g
ˆ
2(v0 cos T0 )2

v 0 2 sin 2 T 0
ˆ The maximum height (H) reached by the projectile, H =
2g

v0 sin T 0
ˆ Time taken to achieve maximum height by the Projectile, tm = g

2 v0 sin T 0
ˆ Time of flight for Projectile, tF = 2tm = g

2 v0 2 sin T 0 cos T 0 v0 2 sin 2 T 0
ˆ Range of a projectile (Distance travelled in horizontal direction) : R = =
g g

v2
ˆ Maximum Range Rmax = 0g (For given v0, q Should be 45° for maximum Range.)

ˆ Relation between range and maximum height for projectile tan q0 = 4RH

ˆ For q and 90° – q, ranges covered by projectiles are equal for given v0.
ˆ The ranges covered by projectile at 45° + a and 45° – a are equal.

(60) Two objects are projected with the same velocity at different angles with the horizontal and if the
range is same for both of them. It t1 and t2 are their time of flights then t1. t2 = ......

(A) 2 Rg (B) 2 Hg (C) (D) 2 g
2R R
g

(61) An object is projected at angle of 30° with the horizontal, with kinetic energy K. Its kinetic
energy at maximum height is ......

(A) 0 (B) 4 (C) (D)
3K K 3K
2 2

(62) For a projectile motion y(t) = 12t – 5t2 and x(t) = 5t. Where x and y are in meter and t in
second. So initial velocity ......
(A) 6 ms–1 (B) 12 ms–1 (C) 5 ms–1 (D) 13 ms–1
(63) Bullets are fired with the same initial velocity v in different direction on a plane surface. These
bullets would fall on the maximum area of ...... on this surface.

Sv 2 S 2v 2 Sv 4 Sv 2
(A) (B) (C) (D)
g g2 g2 g2

(64) In a projectile motion, if the maximum height H is one-forth of the range (R), then angle of
projections with horizontal is ......
(A) zero (B) 30° (C) 45° (D) 60°

Ans. : 60 (C), 61 (B), 62 (D), 63 (C), 64 (C)

31

Graphs :
ˆ Graph is always y ® x.
ˆ Shape of graph can be decided by power of y and x in a equation express in terms of y.
– If both have a power 1-1 then graph is straight line, otherwise different shape for different
power. (generally they are considered as a parabolic).
ˆ For a straight line,
– Co-efficient of x (with sign) gives a slope of graph when equation is expressed in terms of Y.
– To obtain a segment on Y-axis, put x-co-ordinate zero and get the value of y.
– To obtain a segment on X-axis, put y-co-ordinate zero and get the value of x.
ˆ If graph is strait line, then equation of line will be y = mx + c.
Where m = slope of line and c = segment on y axis.
ˆ From a graph, a physical quantity which is obtained by product of x and y-axis will be the area
under the curve and x-axis.
ˆ From a graph, a ratio of y to x is obtained by a slope of curve at a point.

Distance ® time velocity ® time Important-point
graph graph

Uniform distance (d) velocity (v) (1) d = 0 at t = 0, in graph of
motion displacement ® time.
(motion with (2) Slope of graph of displacement ® time
v = constant
d = vt
constant = velocity = constant
velocity) 0
(3) Graph of velocity ® time is parallel
0 time (t) time (t)
to time x-axis, so slope = 0
it means acceleration = 0
Constant distance (d) velocity (v) (1) Graph of displacement ® time is
acceleration Parabolic
motion v = at (2) Slope of velocity ® time graph
(i) initial velocity = acceleration = constant
1 2
d= at
2
and initial 0 (3) Initial velocity and initial position Q is
0 time (t)
time (t)
position is zero zero, So graph starts from origin.

distance (d)
velocity (v) (1) Here initial position is zero, so graph
(ii) Initial of displacement ® time starts
d = v0t + at
1 2
velocity ¹ 0 2 v = v0 + at from origin.
Initial v0 (2) Initial velocity is not zero, so graph
position = 0 time (t) of velocity ® time does not start from origin
0 time (t)
(3) Slope of velocity ® time graph
= acceleration = constant.

32

distance (d) velocity (1) Initial position and Initial velocity are
(v)
(iii) Initial at non zero so, graphs do not pass
t
2
+
velocity ¹ 0 1 a v0 through origin.
t+ 2 =
v0 v0 v
Initial r0 +
r0 (2) Slope of velocity ® time
=
d
position ¹ 0 0 time (t)
0
time (t) = acceleration = constant
constant dece- distance (d) velocity (v)
lerated motion (1) Slope of graph velocity ® time is
(i) Till final 0 v0 constant but negative
=
v
velocity (2) Final velocity becomes zero, So
becomes zero 0 t0 motion will be posible till t = t0.
0
time (t)
t = t0 time (t)

distance (d) velocity (v)
(i) It final (1) Slope of graph velocity ® time is
velocity v0
constant but negative.
becomes (2) Final velocity not become zero,
negative So motion will continue after t = t0
0 t0 time (t)
0 t0 time (t)

(65) Figure shows a graph of velocity ® time for any particle. It is clear from graph that ...... on
particle in part B.
velocity (v) (A) Force act in direction of motion.
(B) Force act in opposite direction of motion.
(C) Force will be zero
D
B (D) nothing is clear about force.
A C

0 ® time (t)
(66) A graph of displacement ® time makes 30° angle with time axis. After 2s it makes 60° angle
with time angle. So, average acceleration during this motion is ......
1
(A) 3 (B) 1 (C) 2 3 (D)
3

(67) Figure shows a velocity ® time graph for a upward going lift. So, at ...... m height lift becomes
stationary.

velocity (v) ms–1 (A) 12
(B) 32
(C) 44
(D) 24
4

0 3 8
time (t) s

Ans. : 65 (C), 66 (D), 67 (B)

33

Assertion - Reason type Question :
Instruction : Read assertion and reason carefully, select proper option from given below.
(a) Both assertion and reason are true and reason explains the assertion.
(b) Both assertion and reason are true but reason does not explain the assertion.
(c) Assertion is true but reason is false.
(d) Assertion is false and reason is true.
(68) Assertion : Speed will be changed without changing velocity of object.
Reason : When velocity of object is zero then acceleration of it may not be zero.
(A) a (B) b (C) c (D) d
(69) Assertion : An object becomes stationary for a moment while the direction of motion will
changed.
Reason : At given instant, If velocity of object is zero then acceleration of object is also zero.
(A) a (B) b (C) c (D) d
(70) Assertion : The v ® t graph perpendicular to time axis is not possible in practice.
Reason : Infinite acceleration can not be realized in practice.
(A) a (B) b (C) c (D) d
(71) Assertion : Magnitude of average velocity is equal to average speed if velocity is constant.
Reason : If velocity is constant, then there is no change in the direction of motion.
(A) a (B) b (C) c (D) d
(72) Assertion : Particle A is moving northwards and partide B eastwards with same speed. Then,
velocity of A with respect to B is in southeast direction.
Reason : Relative velocity between them is zero as their speeds are same.
(A) a (B) b (C) c (D) d
(73) Assertion : In the v ® t diagram shown in figure,
average velocity between the interval t = 0 and v

t = t1 is independent of t1. vm

Reason : Average velocity for given interval is 2 .
vm

(A) a (B) b
(C) c (D) d
t
Match the columns : t=0 t = t1

(74) Find the appropriate option from column 2 for column-1, when motion is described by
d = 3 + 8t – 4t2.
column-1 column-2
(a) Initial acceleration (p) –16 unit (A) a ® s, b ® r, c ® q, d ® p
(b) velocity at end of 3 s (q) 3 unit (B) a ® s, b ® p, c ® q, d ® r
(c) distance travelled in 2 s (r) 7 unit (C) a ® q, b ® r, c ® s, d ® p
(d) displacement at 1 s (s) –8 unit (D) a ® q, b ® r, c ® p, d ® s

Ans. : 68 (A), 69 (C), 70 (A), 71 (A), 72 (C), 73 (A), 74 (B)
ˆ
34