(D) Initial velocity $u = 20 \; m/s$. Final velocity $v = 0$. Distance $S_1 = 500 \; m$.Using the third equation of motion,$v^2 = u^2 - 2aS_1$:$0 = (2

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(D) Initial velocity $u = 20 \; m/s$. Final velocity $v = 0$. Distance $S_1 = 500 \; m$.Using the third equation of motion,$v^2 = u^2 - 2aS_1$:$0 = (20)^2 - 2 \cdot a \cdot 500$$1000a = 400 \Rightarrow a = 0.4 \; m/s^2$.Now,if the brakes are applied at half the distance,$S_2 = 250 \; m$:$v^2 = u^2 - 2aS_2$$v^2 = (20)^2 - 2 \cdot 0.4 \cdot 250$$v^2 = 400 - 200 = 200$$v = \sqrt{200} \; m/s$.Comparing with $\sqrt{x} \; m/s$,we get $x = 200$.

Class 11 Physics Motion in Straight Line Uniformly Accelerated Motion

Medium

For a train engine moving with a speed of $20 \; m/s$,the driver must apply brakes at a distance of $500 \; m$ before the station for the train to come to rest at the station. If the brakes were applied at half of this distance,the train engine would cross the station with a speed of $\sqrt{x} \; m/s$. The value of $x$ is $..............$ (Assuming the same retardation is produced by the brakes).

JEE MAIN 2023 All JEE MAIN

A
$100$
B
$101$
C
$520$
D
$200$

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Motion in Straight Line

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Uniformly Accelerated Motion

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$A$ body $A$ moves with a uniform acceleration $a$ and zero initial velocity. Another body $B$ starts from the same point and moves in the same direction with a constant velocity $v$. The two bodies meet after a time $t$. The value of $t$ is:

Easy

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