Motion in a Straight Line — Page 4 | NEET Physics

Q31

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

A 24 m s

-

2

B zero

C 6 m s

-

2

D 12 m s

-

2

Correct Answer

Option D

Solution

Given x = 8 + 12t $-$ t 3 Velocity, v =

{{dx} \over {dt}}

= 12 - 3t 2 When v = 0, then 12 - 3t 2 = 0 $\Rightarrow$ t = 2 s

a = {{dv} \over {dt}}

= - 6t $\therefore$ At t = 2 s,

a

= - 12 m/s 2 $\therefore$ Retardation = 12 m/s 2

Q32

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

A

{{{v_1} + {v_2}} \over 2}

B

{{{v_1}{v_2}} \over {{v_1} + {v_2}}}

C

{{2{v_1}{v_2}} \over {{v_1} + {v_2}}}

D

{{v_1^2v_2^2} \over {v_1^2 + v_2^2}}

Correct Answer

Option C

Solution

Let total distance = 2S Let particle take t 1 time to cover first S distance, $\therefore$ t 1 =

{S \over {{v_1}}}

Let particle take t 2 time to cover last S distance, $\therefore$ t 2 =

{S \over {{v_2}}}

$\therefore$ Average speed =

{{2S} \over {{S \over {{v_1}}} + {S \over {{v_2}}}}}

=

{{2{v_1}{v_2}} \over {{v_1} + {v_2}}}

Q33

A boy standing at the top of a tower of 20 m height drops a stone. Assuming g = 10 m s

-

2 , the velocity with which it hits the ground is

A 10.0 m/s

B 20.0 m/s

C 40.0 m/s

D 5.0 m/s

Correct Answer

Option B

Solution

We know, v 2 = u 2 + 2gh Here u = 0 $\therefore$ v =

\sqrt {2gh}

=

\sqrt {2 \times 10 \times 20}

= 20 m/s

Q34

A particle moves a distance x in time t according to equation x = (t + 5)

-

1 . The acceleration of particle is proportional to

A (velocity) 3/2

B (distance) 2

C (distance)

-

2

D (velocity) 2/3

Correct Answer

Option A

Solution

x =

{1 \over {t + 5}}

$\therefore$ v =

{{dx} \over {dt}}

=

{{ - 1} \over {{{\left( {t + 5} \right)}^2}}}

.......(i) $\therefore$

a

=

{{dv} \over {dt}}

=

{2 \over {{{\left( {t + 5} \right)}^3}}}

= 2x 3 .......(ii) $\therefore$

a

$\propto$ (distance) 3 From equation (i), we get

{v^{3/2}} = {{ - 1} \over {{{\left( {t + 5} \right)}^3}}}

Putting this in equation (ii), we get

a

= -2

{v^{3/2}}

$\therefore$

a

$\propto$ (velocity) 3/2

Q35

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 = 18 s. What is the value of v? (Take g = 10 m/s 2 )

A 75 m/s

B 55 m/s

C 40 m/s

D 60 m/s

Correct Answer

Option A

Solution

From the question, we can say distance moved by 1 st ball in 18 s = distance moved by 2 nd ball in 12 s.

So, distance moved by 1 st ball in 18 s =

{1 \over 2} \times 10 \times {18^2}

= 1620 m and distance moved by 2 nd ball in 12 s =

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

$\therefore$ 1620 =

v\left( {12} \right) + {1 \over 2} \times 10{\left( {12} \right)^2}

$\Rightarrow$

v

= 75 m/s

Q36

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 chase the bus ?

A 40 m s

-

1

B 25 m s

-

1

C 10 m s

-

1

D 20 m s

-

1

Correct Answer

Option D

Solution

Relative velocity of the scooter with respect to the bus = (v s - 10) $\therefore$

{{1000} \over {{v_s} - 10}}

= 100 $\Rightarrow$ v s = 20 m/s

Q37

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

A S 2 = 3S 1

B S 2 = 4S 1

C S 2 = S 1

D S 2 = 2S 1

Correct Answer

Option B

Solution

Given u = 0. S 1 =

{1 \over 2} \times a \times t_1^2

and S 2 =

{1 \over 2} \times a \times t_2^2

{{{S_1}} \over {{S_2}}} = {\left( {{{{t_1}} \over {{t_2}}}} \right)^2}

=

{\left( {{{10} \over {20}}} \right)^2}

=

{1 \over 4}

S 2 = 4S 1

Q38

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

A 12

B 9

C 10

D 1.8

Correct Answer

Option B

Solution

v 2 = u 2 + 2

a

s $\Rightarrow$

a = {{{v^2} - {u^2}} \over {2s}}

=

{{{{\left( {20} \right)}^2} - {{\left( {10} \right)}^2}} \over {2 \times 135}}

=

{{400 - 100} \over {270}}

=

{{300} \over {270}}

=

{{10} \over 9}

m/s 2

v = u + at

$\Rightarrow$

t = {{v - u} \over a}

=

{{20 - 10} \over {{{10} \over 9}}}

=

{{10} \over {{{10} \over 9}}}

= 9 m/s

Q39

The distance travelled by a particle starting from rest and moving with an acceleration

{4 \over 3}

m s

-

2 , in the third second is

A

{{10} \over 3}

m

B

{{19} \over 3}

m

C 6 m

D 4 m

Correct Answer

Option A

Solution

distance travelled in nth seconds S(n th) = u +

{a \over 2}(2n - 1)

Here u = 0,

a

=

{4 \over 3}

m s $-$ 2 , n = 3 S(3 rd) =

{4 \over {3 \times 2}}(2 \times 3 - 1)

=

{{10} \over 3}

m

Q40

A particle moving along x-axis has acceleration f, at time t, given by f = f 0

\left( {1 - {t \over T}} \right),

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

A

{1 \over 2}

f 0 T 2

B f 0 T 2

C

{1 \over 2}

f 0 T

D f 0 T

Correct Answer

Option C

Solution

f = f 0

\left( {1 - {t \over T}} \right)

When f = 0 then, 0 = f 0

\left( {1 - {t \over T}} \right)

$\Rightarrow$

\left( {1 - {t \over T}} \right)

= 0 $\Rightarrow$ t = T Also f =

{{dv} \over {dt}}

$\therefore$

\int\limits_0^{{v_x}} {dv = \int\limits_{t = 0}^{t = T} {fdt} }

v x =

{\int\limits_{t = 0}^{t = T} {{f_0}\left( {1 - {t \over T}} \right)dt} }

=

{{f_0}\left[ {\left( {t - {{{t^2}} \over {2T}}} \right)} \right]_0^T}

=

{{f_0}\left( {T - {{{T^2}} \over {2T}}} \right)}

=

{{f_0}\left( {T - {T \over 2}} \right)}

=

{1 \over 2}

f 0 T