Question

A car starts from rest and moves with uniform acceleration$a$ on a straight road from time$t=0tot=T$. 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. $aT$

Answer 1

The correct option is C $\frac{aT}{2}$

For First part,

u = 0, t = T and acceleration = a
$\therefore v=0+aT$ 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 $v^2=u^2-2a{S}_2\Rightarrow S_2=\frac{u^2}{2a_1}=\frac{1}{2}\frac{a^2{T}^2}{a_1}$

$S_2=\frac{1}{2}aT\times T_1(As{a}_1=\frac{aT}{T_1})$

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

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