A car starts from rest and moves with uniform | vedclass

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Question

(c) For First part,

$u = 0, t = T $ and acceleration $= a$

$	herefore v = 0 + aT = aT$and ${S_1} = 0 + \frac{1}{2}a{T^2} = \frac{1}{2}a{T^2}$

Vedclass · Accepted Answer

$\frac{{aT}}{2}$

Vedclass · Answer

(c) For First part,

$u = 0, t = T $ and acceleration $= a$

$	herefore v = 0 + aT = aT$and ${S_1} = 0 + \frac{1}{2}a{T^2} = \frac{1}{2}a{T^2}$

For Second part,

$u = aT,$ retardation $=a_1$, $v = 0$ and time taken $= T_1$ (let)

$	herefore $$0 = u - {a_1}{T_1}$$ \Rightarrow aT = {a_1}{T_1}$

and from ${v^2} = {u^2} - 2a{S_2}$ $ \Rightarrow {S_2} = \frac{{{u^2}}}{{2{a_1}}} = \frac{1}{2}\frac{{{a^2}{T^2}}}{{{a_1}}}$

${S_2} = \frac{1}{2}aT 	imes {T_1}$ $\left( {As\,\,{a_1} = \frac{{aT}}{{{T_1}}}} ight)$

$	herefore $ ${v_{av}} = \frac{{{S_1} + {S_2}}}{{T + {T_1}}} = \frac{{\frac{1}{2}a{T^2} + \frac{1}{2}aT 	imes {T_1}}}{{T + {T_1}}}$

$ = \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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