31 Year NEET Previous Year Questions: Motion in a Straight Line - 1
16 Questions MCQ Test Physics Class 11 | 31 Year NEET Previous Year Questions: Motion in a Straight Line - 1
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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]
, 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]
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
The distance travelled by a particle starting from rest and moving with an acceleration 4/3ms-2 , in the third second is:
Distance travelled in the nth second is given by
put u = 0, 
A particle shows distance - time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point: [2008]
The slope of the graph 
is maximum at C and hence the instantaneous velocity is maximum at C.
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]
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
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]
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
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]
u = 0, t1=10s, t2 = 20s
Using the relation, 
Acceleration being the same in two cases,
S2 = 4S1
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)
Clearly distance moved by 1st bal l in 18s = distance moved by 2nd ball in 12s.
Now, distance moved in 18 s by 1st
= 90 x 18 = 1620 m
Distance moved in 12s by 2nd ball
∴ 1620 = 12 v + 5 x 144
⇒ v = 135 – 60 = 75 ms –1
A particle h as in itial velocity 
and has acceleration 
. It's speed after 10 s is:
⇒ ux = 3 units, uy = 4 units
ax = 0.4 units, ay = 0.3 units
A particle moves a distance x in time t according to equation x = (t + 5)-1. The acceleration of particle is proportional to [2010]
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:
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
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]
Here, u = 0
We have, v2 = u2 + 2gh
= 20 m/s
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]
Average acceleration
< a > = 5 m/sec2
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
Let the total distance covered by the particle be 2s. Then
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 :
x = 8 + 12t – t3
The final velocity of the particle will be zero, because it retarded.
V = 0 + 12 – 3t2 = 0
3t2 = 12
t = 2 sec
Now the retardation
a [t = 2] = – 12 m/s2 retardation = 12 m/s2
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]
⇒ h2 = 375
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]
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