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Test: Kinematic Equations for Uniformly Accelerated Motion - NEET MCQ


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30 Questions MCQ Test Physics Class 11 - Test: Kinematic Equations for Uniformly Accelerated Motion

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Test: Kinematic Equations for Uniformly Accelerated Motion - Question 1

A body starts from rest and moves with constant acceleration for t s. It travels a distance x1 in first half of time and x2 in next half of time, then

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 1

As the body starts from rest,
∴ u = 0
Let a be constant acceleration of the body.
Distance travelled by the body in (t/2) s is
x1=ut + 1/2 at2

Distance travelled by the body in t s is

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 2

Free fall of an object in vacuum is a case of motion with

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 2

Free fall of an object in vacuum is a case of motion with uniform acceleration.

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Test: Kinematic Equations for Uniformly Accelerated Motion - Question 3

Which of the following equations does not represent the kinematic equations of motion?

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 3

S = vt + 1/2at2
It is not a kinematic equation of motion.
All others are three kinematic equations of motion.

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 4

Which of the following statements is not correct?

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 4

The sign of acceleration does not tell us whether the particle's speed is increasing or decreasing. The sign of acceleration depends on the choice of the positive direction of the axis.
For example: If the vertically upward direction is chosen to be positive direction of the axis, the acceleration due to gravity is negative. If a particle is falling under gravity, this acceleration though negative results in increase in speed.

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 5

The distances covered by a freely falling body in its first, second, third,..., nth seconds of its motion

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 5

Distance travelled by a body in nth second is

Here, u = 0, a = g
∴ Distance travelled by the body in 1st second is

Distance travelled by the body in 2nd second is

Distance travelled by the body in 3rd second is

and so on.
Hence, the distance covered by a freely falling body in its first, second, third ..... nth second of its motion forms an arithmetic progression.

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 6

A body covers 20 m, 22 m, 24 m, in 8th, 9th and 10th seconds respectively. The body starts

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 6

As distance travelled in successive seconds differ by 2 m each, therefore acceleration is constant = 2ms−2

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 7

A player throws a ball vertically upwards with velocity u. At highest point,

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 7

At the highest point velocity of the ball becomes zero, but its acceleration is equal to g.

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 8

A car moving along a straight road with speed of 144 km/h is brought to a stop within a distance of 200 m. How long does it take for the car to stop?

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 8


Test: Kinematic Equations for Uniformly Accelerated Motion - Question 9

An auto travelling along a straight road increases its speed from 30.0 m s-1 to 50.0 m s-1 in a distance of 180 m. If the acceleration is constant, how much time elapses while the auto moves this distance?

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 9

Let a be constant acceleration of auto.
Here, u = 30ms−1, v = 50ms−1, S = 180 m
As v− u= 2aS


Solving this quadratic equation by quadratic formula, we get = 4.5s, −18 s, (t can't be negative)
∴ t = 4.5 s

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 10

A body falling freely under gravity passes two points 30 m apart in 1 s. From what point above the upper point it began to fall? (Take g = 9.8 ms2).

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 10

Suppose the body passes the upper point at t second and lower point at (t + 1) s, then

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 11

A player throws a ball upwards with an initial speed of 30 m s−1. How long does the ball take to return to the player's hands? (Take g = 10 m s−2).

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 11

Let t be time taken by the ball to reach the highest point.
As v = u + at
Here, u = 30ms−1
v = 0 (At highest point velocity is zero)
a = - g = −10ms−2
∴0 = 30 - 10t or t = 3 s
∴ Time taken by the ball to return to player's hand
= 3 s + 3 s = 6 s.

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 12

A girl standing on a stationary lift (open from above) throws a ball upwards with initial speed 50 m s-1. The time taken by the ball to return to her hands is (Take g = 10 ms-2)

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 12


Test: Kinematic Equations for Uniformly Accelerated Motion - Question 13

A girl standing on a stationary lift (open from above) throws a ball upwards with initial speed 50 m s-1. if the lift starts moving up with a uniform speed of 5ms−1 and the girl again throws the ball up with the same speed, how long does the ball take to return to her hands?

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 13

When the lift starts moving upwards with uniform speed of 5 ms−1, there is no change of relative velocity of ball w.r.t. girl. Hence, even in this case the ball will return to the girl’s hands in the same time i.e 10s.

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 14

It is a common observation that rain clouds can be at about 1 km altitude above the ground. If a rain drop falls from such a height freely under gravity, then what will be its speed in km h-1 (Take g = 10 m )

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 14

Here, u = 0, g = 10 ms-2, h = 1 km = 1000 m
As v- u2 = 2gh
∴ v= 2gh

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 15

A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and other after a time interval (less than 2 s). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is 15 m at t = 2 s. The gap is found to remain constant. The velocities with which the balls were thrown are (Take g = 10 m s−2).

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 15

For first stone,
taking the vertical upwards motion of the first stone up to highest point
Here, u = u1, v = 0 (At highest point velocity is zero)
a = -g, S = h1
As v− u= 2aS

For second stone,
Taking the vertical upwards motion of the second stone up to highest point
here, u = U2, v = 0, a = −g, S = h2
As v− u2 = 2as


As per question

Subtract (ii) from (i), we get,

On substituting the given information, we get

or u= 20ms−1 and u= U1/2 = 10ms-1

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 16

A body sliding on a smooth inclined plane requires 4 seconds to reach the bottom starting from rest at the top. How much time does it take to cover one-fourth distance starting from rest at the top

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 16

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 17

A ball is thrown vertically upwards with a velocity of 20 m s-1 from the top of a multistorey building of 25 m high. How high will the ball rise? (Take g = 10 m s-2)

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 17

Taking vertical upward motion of the ball upto highers point
Here, u = 20 ms-1
v = 0 (at highest point velocity is zero)
a = -g = -10ms-2
As v= u+ 2as
0 = (20)2 + 2(-10)(S) or 

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 18

A ball is thrown vertically upwards with a velocity of 20 m s-1 from the top of a multistorey building of 25 m high. the time taken by the ball to reach the ground is

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 18

Let t1 be the time taken by the ball to reach the highest point.
here, v = 0, u = 20ms−1, a = −g = −10ms−2, t = t1
As v = u + at
∴ 0 = 20 + (−10)t1 or t= 2s
Taking vertical downward motion of the ball from the highest point to ground.
Here, u = 0, a = +g = 10ms−2, S = 20 m + 25 m = 45 m, t = t2


Total time taken by the ball to reach the ground = t+ t= 2s + 3s = 5s

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 19

Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h-1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 ms-2. If after 50s, the guard of B just brushed past A, what was the original distance between them?

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 19


For train B,


Original distance between A and B = SB − SA = 2250m − 1000m = 1250m

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 20

The velocity of a particle at an instant is 10 ms-1 After 3 s its velocity will becomes 16 ms-1. The velocity at 2 s, before the given instant will be

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 20

Here, u = 10ms−1, t = 3s, v = 16ms−1

Now velocity at 2 s, before the given instant
10 = u + 2 x 2
∴ u = 6 m s−1 (since v = u + at)

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 21

A body covers a distance of 4 m in 3rd second and 12 m in 5th second. If the motion is uniformly accelerated, how far will it travel in the next 3 seconds?

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 21


On solving, u = −6ms−1, a = 4ms−2
Distance travelled in next 3 seconds = S− S5

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 22

Stopping distance of a moving vehicle is directly proportional to

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 22

Let ds is the distance travelled by the vehicle before it stops.

Here, final velocity v = 0, initial velocity = u, S = ds

Using equation of motion

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 23

A car moving with a speed of 50 km h-1 can be stopped by brakes after atleast 6 m. If the same car is moving at a speed of 100 km h the minimum stopping distance is

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 23



Test: Kinematic Equations for Uniformly Accelerated Motion - Question 24

An object falling through a fluid is observed to have acceleration given by a = g − bv where g = gravitational acceleration and b is constant. After a long time of release, it is observed to fall with constant speed. The value of constant speed is

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 24

Here, a = g − bv
When an object falls with constant speed vc, its acceleration becomes zero.
∴ g − bv= 0 or v= g/b

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 25

A particle is released from rest from a tower of height 3h. The ratio of the intervals of time to cover three equal heights h is

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 25

Let t1, t2, t3 be the timings for three successive equal heights h covered during the free fall of the particle. Then



Subtracting (i) from (ii), we get

From (i),(iv) and (v), we get
t1:t2:t= 1:(√2-1) : (√3-√2)

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 26

A stone is dropped from the top of a tall cliff and n seconds later another stone is thrown vertically downwards with a velocity u. Then the second stone overtakes the first, below the top of the cliff at a distance given by

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 26

Let the two stones meet at time t.
for the first stone,

for the second stone,


Displacement is same
∴ S1 = S2


Test: Kinematic Equations for Uniformly Accelerated Motion - Question 27

A motorcycle and a car start from rest from the same place at the same time and travel in the same direction. The motorcycle accelerates at 1.0ms−1 up to a speed of 36 kmh-1 and the car at 0.5 ms1 up to a speed of 54 kmh-1. The time at which the car would overtake the motorcycle is

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 27

When car overtakes motorcycle, both have travelled the same distance in the same time. Let the total distance travelled be S and the total time taken to overtake be t.

For motorcycle:
Maximum speed attained = 36kmh−1

Since its acceleration = 1.0ms−2, the time t1 taken by it to attain the maximum speed is given by

The distance covered by motorcycle in attaining the maximum speed is

The time during which the motorcycle moves with maximum speed is (t − 10)s.
The distance covered by the motorcycle during this time is 

∴ Total distance travelled by motorcycle in time t is

For car:
Maximum speed attained =

Since its acceleration = 0.5ms−2

The time taken by it to attain the maximum speed is given by
15 = 0 + 0.5 x t2 or t= 30s (∵ u = 0)
The distance covered by the car in attaining the maximum speed is 

The time during which the car moves with maximum speed is (t − 30)s.
The distance covered by the car during this time is

∴ Total distance travelled by car in time t is

From equations (i) and (ii), we get
10t − 50 = 151 − 225 or 51 = 175 or 1 = 35s

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 28

A body initially at rest is moving with uniform acceleration a. Its velocity after n seconds is v. The displacement of the body in last 2 s is

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 28

Displacement in last 2 second

Acceleration a = v/n
(∵ t = ns)
Displacement in last 2s =

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 29

A body A starts from rest with an acceleration a1. After 2 seconds, another body B starts from rest with an acceleration a2. If they travel equal distances in the 5th second, after the start of A, then the ratio a: a2,  is equal to

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 29

Time taken by body A, t= 5s

Acceleration of body A = a1

Time taken by body B, t= 5 − 2 = 3s

Acceleration of body B = a2

Distance covered by first body in 5th second after its start,


Distance covered by the second body in the 3rd second after its start,

Since S= S3

Test: Kinematic Equations for Uniformly Accelerated Motion - Question 30

A bullet fired into a fixed wooden block loses half of its velocity after penetrating 40cm. It comes to rest after penetrating a further distance of

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion - Question 30

For first part of penetration, by equation of motion

For later part of penetration

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