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QUESTION: 1

A particle moving along x-axis has acceleration f, at time t, given by , where f_{0} 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 (v_{x}) is

[2007]

Solution:

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

QUESTION: 2

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

Solution:

Distance travelled in the nth second is given by

put u = 0,

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]

Solution:

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

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]

Solution:

Initial velocity, u = 10 ms^{–1}

Final velocity, v = 20 ms^{–1 }

Distance, s = 135 m

Let, acceleration = a

Using the formula, v^{2} = u^{2} + 2as

Now, using the relation, v = u + at

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]

Solution:

Let v be the relative velocity of scooter w.r.t bus as v = v_{S}– v_{B}

= 10 + 10 = 20 ms^{–1 }

∴ velocity of scooter = 20 ms^{–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 S_{1} and that covered in the first 20 seconds is S_{2}, then: [2009]

Solution:

u = 0, t_{1}=10s, t_{2} = 20s

Using the relation,

Acceleration being the same in two cases,

S_{2 }= 4S_{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/s^{2})

Solution:

Clearly distance moved by 1^{st }bal l in 18s = distance moved by 2^{nd} ball in 12s.

Now, distance moved in 18 s by 1^{st}

= 90 x 18 = 1620 m

Distance moved in 12s by 2^{nd} ball

∴ 1620 = 12 v + 5 x 144

⇒ v = 135 – 60 = 75 ms ^{–1}

QUESTION: 8

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

Solution:

⇒ u_{x} = 3 units, u_{y} = 4 units

a_{x} = 0.4 units, a_{y} = 0.3 units

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]

Solution:

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:

Solution:

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

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]

Solution:

Here, u = 0

We have, v^{2} = u^{2} + 2gh

= 20 m/s

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]

Solution:

Average acceleration

< a > = 5 m/sec^{2}

QUESTION: 13

A particle covers half of its total distance with speed v_{1} and the rest half distance with speed v_{2}. Its average speed during the complete journey is

Solution:

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

QUESTION: 14

The motion of a particle along a straight line is described by equation : [2012] x = 8 + 12t – t^{3} where x is in metre and t in second. The retardation of the particle when its velocity becomes zero, is :

Solution:

x = 8 + 12t – t^{3}

The final velocity of the particle will be zero, because it retarded.

V = 0 + 12 – 3t^{2} = 0

3t^{2} = 12

t = 2 sec

Now the retardation

a [t = 2] = – 12 m/s^{2} retardation = 12 m/s^{2}

QUESTION: 15

A stone falls freely under gravity. It covers distances h_{1}, h_{2} and h_{3} in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between h_{1}, h_{2} and h_{3} is [NEET 2013]

Solution:

⇒ h_{2} = 375

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]

Solution:

### Motion in straight line

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