Motion in a Straight Line — Page 3 | NEET Physics

Q21

The displacement-time graphs of two moving particles make angles of 30

^\circ

and 45

^\circ

with the x-axis as shown in the figure. The ratio of their respective velocity is

A

\sqrt3

: 1

B 1 : 1

C 1 : 2

D 1 :

\sqrt3

Correct Answer

Option D

Solution

Slope of x-t curves gives the velocity $\Rightarrow$ Ratio

= {{\tan 30^\circ } \over {\tan 45^\circ }} = {1 \over {{{\sqrt 3 } \over 1}}} = 1:\sqrt 3

Q22

A small block slides down on a smooth inclined plane, starting from rest at time t = 0. Let S n be the distance travelled by the block in the interval t = n

-

1 to t = n. The the ratio

{{{S_n}} \over {{S_{n + 1}}}}

is :

A

{{2n} \over {2n - 1}}

B

{{2n - 1} \over {2n}}

C

{{2n - 1} \over {2n + 1}}

D

{{2n + 1} \over {2n - 1}}

Correct Answer

Option C

Solution

{{{S_n}} \over {{S_{n + 1}}}} = {{{a \over 2}(2n - 1)} \over {{a \over 2}\left( {2(n + 1) - 1} \right)}} = {{2n - 1} \over {2n + 2 - 1}} = {{2n - 1} \over {2n + 1}}

Q23

A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of 80 m/s. The height of the tower is : (g = 10 m/s 2 )

A 340 m

B 320 m

C 300 m

D 360 m

Correct Answer

Option C

Solution

From the Kinematic equation v 2 = u 2 + 2gh v = 80 m/s u = 20 m/s h =

{{{v^2} - {u^2}} \over {2g}}

= {{6400 - 400} \over {20}}

= 300 m

Q24

The speed of a swimmer in still water is 20 m/s. The speed of river water is 10 m/s and is flowing due east. If he is standing on the south bank and wishes to cross the river along the shortest path, the angle at which he should make his strokes w.r.t. north is given by :

A 0 o

B 60

C 45 o west

D 30 o west

Correct Answer

Option D

Solution

V SR = 20 m/s V RG = 10 m/s

{\overrightarrow V _{SG}} = {\overrightarrow V _{SR}} + {\overrightarrow V _{RG}}

sin $\theta$ =

\left| {{{{{\overrightarrow V }_{RG}}} \over {{{\overrightarrow V }_{SR}}}}} \right|

=

{{10} \over {20}}

=

{1 \over 2}

$\Rightarrow$ θ = 30° west-side

Q25

Preeti reached the metro station and found that the escalator was not working. She walked up the sationary escalator in time t 1 . On another days, if she remains stationary on the the moving escalator, then the escalator takes her up in time t 2 . The time taken by her to walk up on the moving escalator will be

A

{{{t_1}{t_2}} \over {{t_2} - {t_1}}}

B

{{{t_1}{t_2}} \over {{t_2} + {t_1}}}

C

{{t_1} - {t_2}}

D

{{{t_1} + {t_2}} \over 2}

Correct Answer

Option B

Solution

Velocity of preeti with respect to elevator v 1 =

{d \over {{t_1}}}

Velocity of elevator with respect to ground v 2 =

{d \over {{t_2}}}

$\therefore$ Net velocity of preeti on moving escalator with respect to the ground v = v 1 + v 2

{d \over {{t}}}

=

{d \over {{t_1}}}

+

{d \over {{t_2}}}

{1 \over {{t}}}

=

{1 \over {{t_1}}}

+

{1 \over {{t_2}}}

$\therefore$ t =

{{{t_1}{t_2}} \over {{t_1} + {t_2}}}

Here t is the time taken by preeti to walk up on the moving escalator.

Q26

Two cars P and Q start from a point at the same time in a straight line and their positions are represented by x P (t) = (at + bt 2 ) and x Q (t) = (ft

-

t 2 ). At what time do the cars have the same velocity ?

A

{{a - f} \over {1 + b}}

B

{{a + f} \over {2\left( {b - 1} \right)}}

C

{{a + f} \over {2\left( {1 + b} \right)}}

D

{{f - a} \over {2\left( {1 + b} \right)}}

Correct Answer

Option D

Solution

For car P, x P (t) = (at + bt 2 ) v P (t) =

{{d{x_p}\left( t \right)} \over {dt}}

= a + 2bt Similarly for car Q, x Q (t) = (ft - t 2 ) v Q (t) =

{{d{x_Q}\left( t \right)} \over {dt}}

= f - 2t When they have same velocity then, v P (t) = v Q (t) $\therefore$ a + 2bt = f - 2t $\Rightarrow$ 2t(b + 1) = f - a $\Rightarrow$ t =

{{f - a} \over {2\left( {1 + b} \right)}}

Q27

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

A

{3 \over 2}A + {7 \over 3}B

B

{A \over 2} + {B \over 3}

C

{3 \over 2}A + 4B

D

3A + 7B

Correct Answer

Option A

Solution

Given, v = At + Bt 2

{{dx} \over {dt}}

= At + Bt 2

\int {dx = \int {\left( {At + B{t^2}} \right)} } dt

x =

{{A{t^2}} \over 2} + {{B{t^3}} \over 3} + C

At t = 1, particle is at x(t = 1) =

{A \over 2} + {B \over 3} + C

At t = 2, particle is at x(t = 2) =

2A + {{8B} \over 3} + C

$\therefore$ distance travelled by the particle between 1 s and 2 s is, = x(t = 2) - x(t = 1) = (

2A + {{8B} \over 3} + C

) - (

{A \over 2} + {B \over 3} + C

) =

{3 \over 2}A + {7 \over 3}B

Q28

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to

v\left( x \right) = \beta {x^{ - 2n}}

, where

\beta

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

- 2{\beta ^2}{x^{ - 2n + 1}}

B

- 2n{\beta ^2}{e^{ - 4n + 1}}

C

- 2n{\beta ^2}{x^{ - 2n - 1}}

D

- 2n{\beta ^2}{x^{ - 4n - 1}}

Correct Answer

Option D

Solution

Given

v\left( x \right) = \beta {x^{ - 2n}}

$\therefore$

{{dv} \over {dx}} = - 2n\beta {x^{ - 2n - 1}}

So acceleration of the particle is

a

=

v{{dv} \over {dx}}

= (

\beta {x^{ - 2n}}

) $\times$ (

- 2nb{x^{ - 2n - 1}}

) =

- 2n{\beta ^2}{x^{ - 4n - 1}}

Q29

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 =

\sqrt x + 3

. The displacement of the particle when its velocity is zero, will be

A 4 m

B 0 m (zero)

C 6 m

D 2 m

Correct Answer

Option B

Solution

As t =

\sqrt x + 3

$\Rightarrow$

\sqrt x = t - 3

$\Rightarrow$

x = {\left( {t - 3} \right)^2}

Velocity, v =

{{dx} \over {dt}}

= 2(

t - 3

) Velocity of the particle becomes zero, when 2(

t - 3

) = 0 $\Rightarrow$ t = 3 s At t = 3 s,

x = {\left( {3 - 3} \right)^2}

= 0 m

Q30

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

A h 2 = 3h 1 and h 3 = 3h 2

B h 1 = h 2 = h 3

C h 1 = 2h 2 = 3h 3

D

{h_1}

=

{{{h_2}} \over 3}

=

{{{h_3}} \over 5}

Correct Answer

Option D

Solution

Here h =

{1 \over 2}g{t^2}

$\therefore$ h 1 =

{1 \over 2}g{\left( 5 \right)^2}

= 125 h 1 + h 2 =

{1 \over 2}g{\left( 10 \right)^2}

= 500 $\Rightarrow$ h 2 = 375 h 1 + h 2 + h 3 =

{1 \over 2}g{\left( 15 \right)^2}

= 1125 $\Rightarrow$ h 3 = 625 $\therefore$ h 2 = 3h 1 and h 3 = 5h 1 or

{h_1}

=

{{{h_2}} \over 3}

=

{{{h_3}} \over 5}