A car starts from rest and moves with uniform | vedclass

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(C) For the first part (acceleration phase):Initial velocity $u = 0$,time $t = T$,and acceleration $= a$.Final velocity $v = u + at = 0 + aT = aT$.Dis

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$\frac{aT}{2}$

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(C) For the first part (acceleration phase):Initial velocity $u = 0$,time $t = T$,and acceleration $= a$.Final velocity $v = u + at = 0 + aT = aT$.Distance covered $S_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}aT^2 = \frac{1}{2}aT^2$.For the second part (deceleration phase):Initial velocity $u = aT$,final velocity $v = 0$,and let the time taken be $T_1$.Using $v = u - a_1 T_1$,we get $0 = aT - a_1 T_1$,so $a_1 T_1 = aT$.Distance covered $S_2 = \frac{u^2}{2a_1} = \frac{(aT)^2}{2a_1} = \frac{a^2 T^2}{2(aT/T_1)} = \frac{1}{2}aT T_1$.Average speed $v_{av} = \frac{	ext{Total distance}}{	ext{Total time}} = \frac{S_1 + S_2}{T + T_1}$.$v_{av} = \frac{\frac{1}{2}aT^2 + \frac{1}{2}aT T_1}{T + T_1} = \frac{\frac{1}{2}aT(T + T_1)}{T + T_1} = \frac{1}{2}aT$.

$A$ car starts from rest and moves with uniform acceleration $a$ on a straight road from time $t = 0$ to $t = T$. After that,a constant deceleration brings it to rest. In this process,the average speed of the car is

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