NEET and AIPMT NEET Physics Motion in A Straight Line MCQ Questions Solved

NEET - 2016

Two cars P and Q start from a point at the same time in a straight line and their positions are represented byxpt=at+bt2 and xQ(t)=ft-t2. At what time do the cars have the same velocity?

(a) a-f1+b

(b)a+f2b-1

(c) a+f21+b

(d) f-a21+b

Concept Videos :-

#13-Graphical-Meaning-of-Instantaneous
#14-Solved-Examples-7
#12-Instantaneous-Velocity

Concept Questions :-

Instantaneous speed and instantaneous velocity

(d) Velocity of each car is given by 

              vp=dxtdt=a+2bt

and vQ=dxQtdt=f-2t

It is given that vP=vQ      a+2bt=f-2t           t=f-a2b+1 

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NEET - 2016

If the velocity of a particle is v= At+Bt2, where A and B are constants, then the distance travelled by it between 1s and 2s is 

(a)3A+7B (b)3/2A+7/3B (c)A/2+B/3 (d)3/2A+4B 

Concept Videos :-

#4-Solved Examples 1
#5-Solved-Examples-2
#6-Solved-Examples-3
#7-Solved-Examples-4
#1-Understanding Position
#2-Understanding Path Length
#3-Understanding Displacement

Concept Questions :-

Distance and displacement


(b) Velocity of the particle is given as v=At+Bt where A and B are constants. 
=> dx/dt=At+Bt [v=dx/dt]
=> dx=(At+Bt2)dt

Integrating both sides, we get 
x2∫dx=2∫(At+Bt2)dt =>Δx=x2-x1=A2∫t dt+B 2∫ t2dt
x1         1                                                 1               1

=A[t2/2]2 +B[t3/3]2 
               1                1
=A/2(22-12)+B/3(23-13
 Distance travelled between 1s and 2s is
Δx=A/2X(3)+B/3(7)=3A/2+7B/3

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NEET - 2015

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x)=βx-2n where, β and n are constants and x is the position of the particle. The acceleration of the particle as a function of x, is given by

(a) -2nβ2x-2n-1

(b) -2nβ2x-4n-1

(c) -2β2x-2n+1

(d) 2nβ2e-4n+1

Concept Videos :-

#16-Acceleration
#17-Solved-Examples-9
#18 Solved Examples 10
#19-Solved-Examples-11

Concept Questions :-

Acceleration

v=βx-2na=vdvdx=βx-2n×β×-2n×x-2n-1=-2nβ2x-4n-1

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NEET - 2013

A stone falls 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

(a) h1=2h2=3h3

(b) h1=h2/3=h3/5

(c) h2=3h1 and h3=3h2

(d) h1=h2=h3

Concept Videos :-

#26-Equations-for-Uniform-Acceleration
#27-Equations for Uniform Acceleration 28Calculus Method29
#28-Sign Convention
#29-Solved Examples 16
#31-Solved-Examples-18
#32-Solved Examples 19
#33-Solved Examples 20
#34-Solved-Examples-21
#35 Solved Examples 22
#30-Solved Examples 17

Concept Questions :-

Uniformly accelerated motion

(b)

 

                       

Initial velocity u=0Distance travelled in first 5 sec, h1    h1=0+12g52     h1=25 g2               ...........iDistance travelled in first 10 sec, s2    s2=0+12g 102  =100 g2Distance h2=s2-h1                     =75 g2              .................iiDistance h3= s3-s2                     = 12g 152-100g2   =125g2            ...............iii

So,  h1: h2 : h3 = 1:3:5

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NEET - 2012

The motion of a particle along a straight line is described by equation

                        x=8+12t-t3

where, x is in metre and t in second. The retardation of the particle when its velocity becomes zero,is

(a)24 ms-2                                       (b)zero

(c)6 ms-2                                         (d)12 ms-2

Concept Videos :-

#42 Non Uniform Acceleration
#43 Solved Examples 29
#44 Solved Examples 30

Concept Questions :-

Non-uniform acceleration

Given, x=8+12t-t3

We know v=dxdt

and     a=dvdt

So,     v=12-3t2

and      a=-6t

At t=2s

            v=0 and  a=-6t

At         t=2s

                  a=-12m/s2

So, retardation of the particle = 12m/s2.

 

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NEET - 2011

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

(a) 20 m/s                                         (b) 40 m/s

(c) 5 m/s                                           (d) 10 m/s

Concept Videos :-

#26-Equations-for-Uniform-Acceleration
#27-Equations for Uniform Acceleration 28Calculus Method29
#28-Sign Convention
#29-Solved Examples 16
#31-Solved-Examples-18
#32-Solved Examples 19
#33-Solved Examples 20
#34-Solved-Examples-21
#35 Solved Examples 22
#30-Solved Examples 17

Concept Questions :-

Uniformly accelerated motion

Given, g=10 ms-2   and h=20 m

We have 

               v=2gh  =2×10×20=400=20 ms-1 

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NEET - 2011

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 

(a) v1v2v1+v2                                         (b) 2v1v2v1+v2

(c) 2v12v22v12+v22                                         (d)  v1+v22

Concept Videos :-

#8-Average-Speed-26-Average-Velocity
#9-Solved-Examples-5
#10-Solved-Examples-6
#11-Examples-on-Application-of-Graphs
#15-Solved-Examples-8

Concept Questions :-

Average speed and average velocity

Velocity v=sls=vt

 The average speed of particle 

        vav=s+ssv1+sv2vav=2v1v2v1+v2

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NEET - 2010

A ball is dropped from a high rise platform at t=0 starting from rest. After 6s another ball is thrown downwards from the same platform with a speed v. The two balls meet at t=18 s. What is the value of v? (take g=10 ms-2)

(a) 75 ms-2                                   (b) 55 ms-1

(c) 40 ms-1                                   (d) 60 ms-1

Concept Videos :-

#45-Relative motion
#46-Relative motion-1
#47-Relative motion-2
#48-Relative motion-3
#49-Relative motion-4
#50-Relative motion-5

Concept Questions :-

Relative motion in 1-D

For first ball, u=0

            s1=12gt12=12×g182

For second ball, initial velocity =v

           s2=vt2+12gt2

           t2=18-6=12s

        s2=v×12+12g122

Here,           s1=s2

            12g182=12v+12g122

          v=75ms-1

 

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NEET - 2010

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

(a) velocity3/2                                          (b) distance2

(c) distance-2                                          (d) velocity2/3

Concept Videos :-

#16-Acceleration
#17-Solved-Examples-9
#18 Solved Examples 10
#19-Solved-Examples-11

Concept Questions :-

Acceleration

Given, distance x=t+5-1                  ...(i)

Differentiating Eq. (i) w.r.t. t, we get

              dxdt=v=-1t+52               ...(ii)

Again, differentiating Eq.(i)w.r.t. t, we get 

               d2xdt2=a=2t+53         ...(iii)

Comparing Eqs.(ii) and(iii),we get 

                av3/2

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NEET - 2009

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 minimum speed should the scooterist chase the bus?

(a) 20 ms-1                                         (b) 40 ms-1

(c) 25 ms-1                                         (d) 10 ms-1

Concept Videos :-

#45-Relative motion
#46-Relative motion-1
#47-Relative motion-2
#48-Relative motion-3
#49-Relative motion-4
#50-Relative motion-5

Concept Questions :-

Relative motion in 1-D

Let v be the relative velocity of scooter(s) w.r.t bus (B), then

                                     v=vS-vB 

                       

                                vS=v+vB                                ...(i)

Relative velocity=displacement/time

                       =1000100   =10 ms-1

Now, substituting the value of v in Eq. (i), we get

                   vS=10+10=20 ms-1

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