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31 Year NEET Previous Questions: Motion in a Straight Line - 1 Free MCQ


MCQ Practice Test & Solutions: 31 Year NEET Previous Year Questions: Motion in a Straight Line - 1 (16 Questions)

You can prepare effectively for NEET Physics Class 11 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "31 Year NEET Previous Year Questions: Motion in a Straight Line - 1 ". These 16 questions have been designed by the experts with the latest curriculum of NEET 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 30 minutes
  • - Number of Questions: 16

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31 Year NEET Previous Year Questions: Motion in a Straight Line - 1 - Question 1

A particle moving along x-axis has acceleration f, at time t, given by , where f0 and T are constants. The particle at t = 0 has zero velocity. In the time interval between t = 0 and the instant when f = 0, the particle’s velocity (vx) is
[2007]

Detailed Solution: Question 1

Here,

where C is the constant of integration.
At t = 0, v = 0.

If f = 0, then

Hence, particle's velocity in the time interval t = 0 and t = T is given by

31 Year NEET Previous Year Questions: Motion in a Straight Line - 1 - Question 2

The distance travelled by a particle starting from rest and moving with an acceleration 4/3ms-2 , in the third second is:

Detailed Solution: Question 2

31 Year NEET Previous Year Questions: Motion in a Straight Line - 1 - Question 3

A particle shows distance - time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point:                              [2008]

Detailed Solution: Question 3

The slope of the graph   is maximum at C and hence the instantaneous velocity is maximum at C.

31 Year NEET Previous Year Questions: Motion in a Straight Line - 1 - Question 4

A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 ms–1 to 20 ms–1 while passing through a distance 135 m in t second. The value of t is: [2008]

Detailed Solution: Question 4

Initial velocity, u = 10 ms–1
Final velocity, v = 20 ms–1
Distance, s = 135 m
Let, acceleration = a
Using the formula, v2 = u2 + 2as

Now, using the relation, v = u + at

31 Year NEET Previous Year Questions: Motion in a Straight Line - 1 - Question 5

A bus is moving with a speed of 10 ms–1 on a straight road. A scooterist wishes to overtake the bus in 100 s. If the bus is at a distance of 1 km from the scooterist, with what speed should the scooterist ch ase the bus? [2009]

Detailed Solution: Question 5

Let v be the relative velocity of scooter w.r.t bus as v = vS– vB

=   10 + 10 = 20 ms–1

∴ velocity of scooter = 20 ms–1

31 Year NEET Previous Year Questions: Motion in a Straight Line - 1 - Question 6

A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 seconds is S1 and that covered in the first 20 seconds is S2, then: [2009]

Detailed Solution: Question 6

u = 0, t1=10s, t2 = 20s

Using the relation,

Acceleration being the same in two cases,

  S2 = 4S1

31 Year NEET Previous Year Questions: Motion in a Straight Line - 1 - Question 7

A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed v. The two balls meet at t = 18s.What is the value of v? [2010] (take g = 10 m/s2)

Detailed Solution: Question 7

Let the two balls meet at 18 seconds after the first ball is dropped. The first ball is dropped from rest and falls freely for 18 seconds.
Displacement of first ball, S1 = (1/2) × g × (18)2 = (1/2) × 10 × 324 = 1620 m.
The second ball is thrown downwards after 6 seconds, so it moves for (18 - 6) = 12 seconds.
Displacement of second ball, S2 = v × 12 + (1/2) × g × (12)2 = 12v + (1/2) × 10 × 144 = 12v + 720 m.
Since both balls meet at the same point, their displacements are equal:
1620 = 12v + 720
1620 - 720 = 12v
900 = 12v
v = 900 / 12 = 75 m/s.
Therefore, the correct value of v is 75 m/s.

31 Year NEET Previous Year Questions: Motion in a Straight Line - 1 - Question 8

A particle h as in itial velocity and has acceleration . It's speed after 10 s is:

Detailed Solution: Question 8

⇒ ux = 3 units, uy = 4 units

ax = 0.4 units, ay = 0.3 units

31 Year NEET Previous Year Questions: Motion in a Straight Line - 1 - Question 9

A particle moves a distance x in time t according to equation x = (t + 5)-1. The acceleration of particle is proportional to [2010]

Detailed Solution: Question 9

31 Year NEET Previous Year Questions: Motion in a Straight Line - 1 - Question 10

A man of 50 kg mass is standing in a gravity free space at a height of 10 m  above the floor. He throws a stone of 0.5 kg  mass downwards with a speed 2 m/s. When the stone reaches the floor, the distance of the man above the floor will be:

Detailed Solution: Question 10

No external force is acting, there fore, momentum is conserved.
By momentum conservation, 50u + 0.5 × 2 = 0 where u is the velocity of man.

Negative sign of u shows that man moves upward.
Time taken by the stone to reach the ground

∴ when the stone reaches the floor, the distance of the man above floor = 10.1 m

31 Year NEET Previous Year Questions: Motion in a Straight Line - 1 - Question 11

A boy standing at the top of a tower of 20m height drops a stone. Assuming g = 10 ms–2, the velocity with which it hits the ground is [2011]

Detailed Solution: Question 11

Here, u = 0

We have, v2 = u2 + 2gh

= 20 m/s

31 Year NEET Previous Year Questions: Motion in a Straight Line - 1 - Question 12

A body is moving with velocity 30 m/s towards east. After 10 seconds its velocity becomes 40 m/s towards north. The average acceleration of the body is [2011]

Detailed Solution: Question 12

Average acceleration

< a > = 5 m/sec2

31 Year NEET Previous Year Questions: Motion in a Straight Line - 1 - Question 13

A particle covers half of its total distance with speed v1 and the rest half distance with speed v2. Its average speed during the complete journey is 

Detailed Solution: Question 13

Let the total distance covered by the particle be 2s. Then

31 Year NEET Previous Year Questions: Motion in a Straight Line - 1 - Question 14

The motion of a particle along a straight line is described by equation : [2012] x = 8 + 12t – t3 where x is in metre and t in second. The retardation of the particle when its velocity becomes zero, is :

Detailed Solution: Question 14

The position of the particle is given by x = 8 + 12t - t3.

To find the retardation when the velocity becomes zero, differentiate x to get the velocity:

v = dx/dt = 12 - 3t2.

Setting v = 0 gives 12 = 3t2 or t2 = 4, so t = 2 seconds.

The acceleration is the derivative of velocity: a = dv/dt = -6t.

Substituting t = 2, we get a = -12 ms-2

31 Year NEET Previous Year Questions: Motion in a Straight Line - 1 - Question 15

A stone falls freely  under gravity. It covers distances h1, h2 and h3 in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between h1, h2 and h3 is [NEET 2013]

Detailed Solution: Question 15

We are given the equation of motion for free fall:

h = (1/2) * g * t²

Now let's calculate the individual heights:

h₁ = (1/2) * g * (5)² = 125

h₁ + h₂ = (1/2) * g * (10)² = 500

So,
h₂ = 500 - 125 = 375

Next,
h₁ + h₂ + h₃ = (1/2) * g * (15)² = 1125

Therefore,
h₃ = 1125 - (h₁ + h₂) = 1125 - 500 = 625

Now, expressing h₂ and h₃ in terms of h₁:

Since
h₁ = 125
h₂ = 375 = 3 * h₁
h₃ = 625 = 5 * h₁

So we can also write:

h₁ = h₂ / 3 = h₃ / 5

31 Year NEET Previous Year Questions: Motion in a Straight Line - 1 - Question 16

The displacement ‘x’ (in meter) of a particle of mass ‘m’ (in kg) moving in one dimension under the action of a force, is related to time ‘t’ (in sec) by t = . The displacement of the particle when its velocity is zero, will be [NEET Kar. 2013]

Detailed Solution: Question 16

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