Motion in A Straight Line Test - 10

For maximum and minimum displacement, we have the magnitude and direction of maximum and minimum velocity.

As maximum velocity in positive direction is \(v_0\);

maximum velocity in opposite direction(Or minimum velocity) is also \(v_0\).

So, we can say;

Maximum displacement in one direction = \(v_0T\)

Maximum displacement in opposite direction = \(-v_0T \)

Hence, \(- v_0T<x<v_0T\) is the correct statement.

⇒ option(b) is correct.

Time \(t_1\) taken in half distance = \(t_1\) = \(L\over v_1\)

Time \(t_2\) taken in half distance \(t_2\)= \(L\over v_2\)

Total time (t) taken in distance (L+L) = \({L\over v_1}+{L\over v_2}={L(v_2+v_1)\over v_1v_2}\)

Total distance= L+L=2L

\(\therefore\) Average speed \(v_{av}\) = \({Total\ distance\over Total\ time}={{2L\over L(v_2+v_1)}\over v_1v_2}={2v_1v_2\over (v_1+v_2)}\)

Let L be the length of the escalator.

Velocity of girl w.r.t. ground \(v_g={L\over t_1}\)

Velocity of escalator w.r.t. ground \(v_e={L\over t_2}\)

Effective Velocity of girl on moving escalator with respect to ground = \(v_g+v_e={L\over t_1}+{L\over t_2}=L[{1\over t_1}+{1\over t_2}]\)

\(v_{ge}=L[{t_1+t_2\over t_1t_2}]\)

\(\therefore\) Time t taken by girl on moving escalator in going up the distance L is

\(t={distance\over speed}={L\over L({t_1+t_2\over t_1t_2})}={t_1t_2\over t_1+t_2}\)

Hence, verifies the option (c).

x = t - sin t

V = \(dx\over dt \) = 1 - cos t

⇒ v = (1 - cos t)

A = \(dv\over dt\) = \(d(1-cos\ t)\over dt\) = +sin t

a = sin t

For  \(v_{max}\) at cos t minimum i.e., cos t = -1

\(\therefore\) \(v_{max}\) = 1-(-1) = 2

For \(v_{min}\) at cos t maximum i.e., cos t = 1

\(v_{min}\) = 1-1 = 0

Hence, v lies between 0 to 2. Verifies the option (d).

X = t – sin t

Sin t varies between 1 and -1 for t > 0.

x will be always positive x(t) > 0. Verifies answer(a).

v = 1 - cos t

when t = 0;v = 0

when t = \(\pi \over 2\); v = 1

when t = \(\pi\); v = 2

when t =2\(\pi\) ;v = 0

so, for t = 2\(\pi\), so at t > 0, v = 0 not > 0

\(\therefore\) It discards option(b)

a = sin t

when t = 0;a = 0

when t = \(\pi \over 2\);a = -1

when t = \(\pi\);a = 0

when t =2\(\pi\) ;a = -1

\(\therefore\) sin t varies from -1 to 1.

So a will varies from -1 to 1 or a can be (-). So discards option (c). Hence, option (a) and (d) are correct

Motion of ball with respect to observer and relative velocity of body .

As the motion is observed from ground, time to strike ball with walls will be after every 10 seconds.As, the ball is moving with very small speed in the moving train,the direction of ball is same as that of train.Hence direction of motion of ball does not change with respect to observer on Earth But, speed of ball changes after collision so option (a) is incorrect and (b) is correct.

As speed of ball is uniform so average speed at any time remain same or 1m/s with respect to train or ground. So option (c) is correct.

Speed of ball changes when it strike to wall initial speed of ball in the direction of moving train with respect to ground = \(V_{TG}\)=10+1=11m/s.

Speed of ball after collision with side of train = \(V_{BG}\) (opposite to the direction of train) = 10-1 = 9m/s.

\(\therefore\) Change in velocity on collision will be in magnitude = 11-9 = 2m/s. So magnitude of acceleration on both walls of compartment is same but direction will be opposite.