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Test: Kinematic Equations for Uniformly Accelerated Motion (April 23) - NEET MCQ


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10 Questions MCQ Test - Test: Kinematic Equations for Uniformly Accelerated Motion (April 23)

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Test: Kinematic Equations for Uniformly Accelerated Motion (April 23) - Question 1

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

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion (April 23) - Question 1

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 (April 23) - Question 2

Which of the following statements is not correct?

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion (April 23) - Question 2

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.

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Test: Kinematic Equations for Uniformly Accelerated Motion (April 23) - Question 3

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 (April 23) - Question 3

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

Test: Kinematic Equations for Uniformly Accelerated Motion (April 23) - Question 4

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 (April 23) - Question 4

there will be acceleration dur to friction


Test: Kinematic Equations for Uniformly Accelerated Motion (April 23) - Question 5

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 (April 23) - Question 5

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 (April 23) - Question 6

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 m s−2).

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion (April 23) - Question 6

Test: Kinematic Equations for Uniformly Accelerated Motion (April 23) - Question 7

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 (April 23) - Question 7

Test: Kinematic Equations for Uniformly Accelerated Motion (April 23) - Question 8

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 (April 23) - Question 8


For train B,


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

Test: Kinematic Equations for Uniformly Accelerated Motion (April 23) - Question 9

Stopping distance of a moving vehicle is directly proportional to

Detailed Solution for Test: Kinematic Equations for Uniformly Accelerated Motion (April 23) - Question 9

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 (April 23) - Question 10

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 (April 23) - Question 10

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)

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