A car starts from rest and moves with unif

Your Ultimate Guide to JEE & NEET Question Solutions

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

A

$\frac{{aT}}{4}$

B

$\frac{{3aT}}{2}$

C

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

D

$aT$

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

For First part,

$u = 0, t = T $ and acceleration $= a$

$\therefore v = 0 + aT = aT$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)

$\therefore $$0 = u - {a_1}{T_1}$$ \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}}}$

${S_2} = \frac{1}{2}aT \times {T_1}$ $\left( {As\,\,{a_1} = \frac{{aT}}{{{T_1}}}} \right)$

$\therefore $ ${v_{av}} = \frac{{{S_1} + {S_2}}}{{T + {T_1}}} = \frac{{\frac{1}{2}a{T^2} + \frac{1}{2}aT \times {T_1}}}{{T + {T_1}}}$

$ = \frac{{\frac{1}{2}aT\;(T + {T_1})}}{{T + {T_1}}}$ $ = \frac{1}{2}aT$