A train moves from rest with a uniform accele | vedclass

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(B) Let $t_1$ be the time taken to accelerate to maximum speed $v$ and $t_2$ be the time taken to retard to rest.Using $v = u + at$,we get $v = 0 + at

Vedclass · Accepted Answer

$2\sqrt{\frac{s}{a}}, \sqrt{as}$

Vedclass · Answer

(B) Let $t_1$ be the time taken to accelerate to maximum speed $v$ and $t_2$ be the time taken to retard to rest.Using $v = u + at$,we get $v = 0 + at_1 \implies t_1 = \frac{v}{a}$.For retardation,$0 = v - at_2 \implies t_2 = \frac{v}{a}$.Total time $T = t_1 + t_2 = \frac{2v}{a}$.Distance covered during acceleration $s_1 = \frac{1}{2}at_1^2 = \frac{1}{2}a(\frac{v}{a})^2 = \frac{v^2}{2a}$.Distance covered during retardation $s_2 = \frac{v^2}{2a}$.Total distance $s = s_1 + s_2 = \frac{v^2}{a} \implies v^2 = as \implies v = \sqrt{as}$.Substituting $v$ in $T = \frac{2v}{a}$,we get $T = \frac{2\sqrt{as}}{a} = 2\sqrt{\frac{s}{a}}$.Thus,the total time is $2\sqrt{\frac{s}{a}}$ and maximum speed is $\sqrt{as}$.

$A$ train moves from rest with a uniform acceleration $a$. After attaining a maximum speed $v$,it starts moving with uniform retardation $a$. Assuming $s$ is the total distance covered in the unidirectional motion of the train,find its total time of journey and maximum speed.

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