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(B) Let the acceleration be $a$. The point starts from rest at $A$ and moves with acceleration $a$ for time $t$ to reach point $B$.Velocity at $B$ is

Vedclass · Accepted Answer

$(2 + \sqrt{2})t$

Vedclass · Answer

(B) Let the acceleration be $a$. The point starts from rest at $A$ and moves with acceleration $a$ for time $t$ to reach point $B$.Velocity at $B$ is $v = at$.Distance $S_{AB} = \frac{1}{2}at^2$.At $B$,the acceleration becomes $-a$. The point continues to move until its velocity becomes zero at point $C$.Using $v_C = v_B + (-a)t'$,where $v_C = 0$ and $v_B = at$,we get $0 = at - at'$,so $t' = t$.Distance $S_{BC} = v_B t' - \frac{1}{2}at'^2 = (at)t - \frac{1}{2}at^2 = \frac{1}{2}at^2$.Total distance $S_{AC} = S_{AB} + S_{BC} = \frac{1}{2}at^2 + \frac{1}{2}at^2 = at^2$.Total time to reach $C$ is $t + t = 2t$.Now,the point returns from $C$ to $A$ under constant acceleration $a$ starting from rest at $C$.$S_{AC} = \frac{1}{2}at_1^2$,where $t_1$ is the time taken to return.$at^2 = \frac{1}{2}at_1^2 \implies t_1^2 = 2t^2 \implies t_1 = \sqrt{2}t$.Total time $T = 2t + t_1 = 2t + \sqrt{2}t = (2 + \sqrt{2})t$.

$A$ point starts moving in a straight line with a certain acceleration. At a time $t$ after the beginning of motion,the acceleration suddenly becomes a retardation of the same value. The time in which the point returns to the initial point is:

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