(A) Since both trains are moving in the same direction,the initial relative velocity of the express train with respect to the other train is $u_{rel}

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(A) Since both trains are moving in the same direction,the initial relative velocity of the express train with respect to the other train is $u_{rel} = v_1 - v_2$.To avoid a collision,the express train must reduce its relative velocity to zero at the point where it would have collided.Using the first equation of motion for relative velocity: $v_{rel} = u_{rel} - at$.Setting the final relative velocity $v_{rel} = 0$,we get:$0 = (v_1 - v_2) - at$$at = v_1 - v_2$$t = \frac{v_1 - v_2}{a}$

Class 11 Physics Motion in a Straight Line Relative Motion in One Dimension

Medium

An express train is moving with a velocity $v_1$. Its driver finds another train moving on the same track in the same direction with velocity $v_2$. To escape a collision,the driver applies a retardation $a$ to the train. The minimum time to escape a collision will be:

A
$t = \frac{v_1 - v_2}{a}$
B
$t = \frac{v_1^2 - v_2^2}{2a}$
C
None
D
Both

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Motion in a Straight Line

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Relative Motion in One Dimension

$A$ train is moving at a constant speed $V$. When its driver observes another train in front of him on the same track and moving in the same direction with constant speed $u$. If the distance between the trains is $x$,what should be the minimum retardation of the train to avoid a collision?

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$A$ bus is moving with a velocity $10 \, m/s$ 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 velocity should the scooterist chase the bus? (in $m/s$)

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$A$ thief is running away on a straight road in a jeep moving with a speed of $9 \, m/s$. $A$ policeman chases him on a motorcycle moving at a speed of $10 \, m/s$. If the instantaneous separation between the jeep and the motorcycle is $100 \, m$,how long will it take for the policeman to catch the thief?

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$A$ body is projected vertically upward with speed $10 \, m/s$ and another at the same time with the same speed in the downward direction from the top of a tower. The magnitude of the acceleration of the first body with respect to the second is ...... $m/s^2$ (take $g = 10 \, m/s^2$).

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An express train is moving with a velocity $v_1$. Its driver finds another train moving on the same track in the same direction with velocity $v_2$. To escape a collision,the driver applies a retardation $a$ to the train. The minimum time to escape a collision will be:

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$A$ train is moving at a constant speed $V$. When its driver observes another train in front of him on the same track and moving in the same direction with constant speed $u$. If the distance between the trains is $x$,what should be the minimum retardation of the train to avoid a collision?

$A$ car travelling at $15 \,m/s$ overtakes another car travelling at $10 \,m/s$. Assuming each car is $4 \,m$ long, what is the time taken during the overtake (in $\,s$)?

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$A$ body is projected vertically upward with speed $10 \, m/s$ and another at the same time with the same speed in the downward direction from the top of a tower. The magnitude of the acceleration of the first body with respect to the second is ...... $m/s^2$ (take $g = 10 \, m/s^2$).

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