Motion in a Straight Line — Page 5 | NEET Physics

Q41

A car moves from X to Y with a uniform speed v u and returns to Y with a uniform speed v d . The average speed for this round trip is

A

\sqrt {{v_u}{v_d}}

B

{{{v_d}{v_u}} \over {{v_d} + {v_u}}}

C

{{{v_u} + {v_d}} \over 2}

D

{{2{v_d}{v_u}} \over {{v_d} + {v_u}}}

Correct Answer

Option D

Solution

Let total distance = 2S Let car take t 1 time to cover first S distance from X to Y, $\therefore$ t 1 =

{S \over {{v_u}}}

Let car take t 2 time to cover S distance,from Y to X, $\therefore$ t 2 =

{S \over {{v_d}}}

$\therefore$ Average speed =

{{2S} \over {{S \over {{v_u}}} + {S \over {{v_d}}}}}

=

{{2{v_u}{v_d}} \over {{v_u} + {v_d}}}

Q42

The positions x of a particle with respect to time t along x-axis is given by x = 9t 2

-

t 3 where x is in metres and t in seconds. What will be the position of this particle when it achieves maximum speed along the + x direction ?

A 54 m

B 81 m

C 24 m

D 32 m

Correct Answer

Option A

Solution

Speed v =

{{dx} \over {dt}}

=

{d \over {dt}}\left( {9{t^2} - {t^3}} \right)

= 18t - 3t 2 At maximum speed

{{dv} \over {dt}}

= 0 $\Rightarrow$ 18 - 6t = 0 $\Rightarrow$ t = 3 $\therefore$ Position of the particle = 81 - 27 = 54 m

Q43

A particle moves along a straight line OX. At a time t (in seconds) the distance x (in metres) of the particle from O is given by x = 40 + 12t

-

t 3 . How long would the particle travel before coming to rest ?

A 16 m

B 24 m

C 40 m

D 56 m

Correct Answer

Option A

Solution

Given x = 40 + 12t $-$ t 3 v =

{{dx} \over {dt}}

= 12 - 3t 2 When particle comes to rest then v = 0. $\therefore$ 12 - 3t 2 = 0 $\Rightarrow$ t = 2 s Distance travelled before coming to rest, s =

\int\limits_0^2 {vdt}

=

\int\limits_0^2 {\left( {12 - 3{t^2}} \right)dt}

=

{\left[ {12t - {{3{t^3}} \over 3}} \right]_0^2}

= 24 - 8 = 16 m

Q44

A car runs at a constant speed on a circular track of radius 100 m, taking 62.8 seconds for every circular lap. The average velocity and average speed for each circular lap respectively is

A 10 m/s, 0

B 0, 0

C 0, 10 m/s

D 10 m/s, 10 m/s.

Correct Answer

Option C

Solution

Distance covered in one circular loop = 2 $\pi$ r = 2 $\times$ 3.14 $\times$ 100 = 628 m Average speed =

{{628} \over {62.8}} = 10

m/s Net displacement in one loop = 0 $\therefore$ Average velocity =

{0 \over {time}}

= 0

Q45

Two bodies A (of mass 1 kg) and B (of mass 3 kg) are dropped from heights of 16 m and 25 m, respectively. The ratio of the time taken by them to reach the ground is

A 4/5

B 5/4

C 12/5

D 5/12

Correct Answer

Option A

Solution

Here h =

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

For body A, 16 =

{1 \over 2}gt_1^2

For body B, 25 =

{1 \over 2}gt_2^2

$\therefore$

{{16} \over {25}} = {{t_1^2} \over {t_2^2}}

$\Rightarrow$

{{{t_1}} \over {{t_2}}} = {4 \over 5}

Q46

The displacement x of a particle varies with time t as x = ae

-

at + be

\beta

t , where a, b,

\alpha

and

\beta

are positive constants. The velocity of the particle will

A be independent of

\beta

B drop to zero when

\alpha

=

\beta

C go on decreasing with time

D go on increasing with time.

Correct Answer

Option D

Solution

x = ae $-$ $\alpha$ t + be $\beta$ t

{{dx} \over {dt}}

= - $\alpha$ ae $-$ $\alpha$ t + b $\beta$ e $\beta$ t v = - $\alpha$ ae $-$ $\alpha$ t + b $\beta$ e $\beta$ t Velocity will go on increasing with time.

Q47

A ball is thrown vertically upward. It has a speed of 10 m/sec when it has reached one half of its maximum height. How high does the ball rise? (Take g = 10 m/s 2 .)

A 10 m

B 5 m

C 15 m

D 20 m

Correct Answer

Option A

Solution

At height

{h \over 2}

, speed is = 10 m/s and at height h, speed = 0 Using formula, v 2 = u 2 - 2gh 0 = (10) 2 - 2(10)

{h \over 2}

$\Rightarrow$ h =

{{100} \over {10}}

= 10 m

Q48

If a ball is thrown vertically upwards with speed u, the distance covered during the last t seconds of its ascent is

A ut

B

{1 \over 2}

gt 2

C ut

-

{1 \over 2}

gt 2

D (u + gt) t.

Correct Answer

Option B

Solution

Let the time taken by the ball to reach the top most point T. $\therefore$ v = u - gT v = 0 at top most point. T =

{u \over g}

Velocity of the ball after (T - t) second v = u - g(T - t) = u - gT + gt = u - g

{u \over g}

+ gt $\therefore$ v = gt Distance covered during the last t seconds of its ascent is s = (gt)t -

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

=

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

Q49

A man throws balls with the same speed vertically upwards one after the other at an interval of 2 seconds. What should be the speed of the throw so that more than two balls are in the sky at any time ? (Given g = 9.8 m/s 2 )

A more than 19.6 m/s

B at least 9.8 m/s

C any speed less than 19.6 m/s

D only with speed 19.6 m/s.

Correct Answer

Option A

Solution

When more than two balls means minimum three balls are in the sky then time of flight for the first ball should be more than 4 s.

T > 4 s

{{2u} \over g}

> 4 $\Rightarrow$ u > 19.6 m/s

Q50

A particle is thrown vertically upward. Its velocity at half of the height is 10 m/s, then the maximum height attained by it (g = 10 m/s 2 )

A 8 m

B 20 m

C 10 m

D 16 m

Correct Answer

Option C

Solution

At height

{h \over 2}

, speed is = 10 m/s and at height h, speed = 0 Using formula, v 2 = u 2 - 2gh 0 = (10) 2 - 2(10)

{h \over 2}

$\Rightarrow$ h =

{{100} \over {10}}

= 10 m