Question

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. $\frac{aT}{2}$
• C. $\frac{3aT}{4}$
• D. $aT$

Answer 1

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

Step: 1 Find distance travelled by the car when it is accelerating.
Formula Used: $S=ut+\frac{1}{2}a{t}^2$
As car starts from rest,
therefore $u=0,t=T$ and acceleration $=a$

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

$S_1=0+\frac{1}{2}a{T}^2=\frac{1}{2}a{t}^2$

Step: 2 Find distance travelled by the car when it is decelerating.
In this situation car will stop, therefore $v=0,u=aT$, retardation $=a_1$, and time taken $=T_1$ (let)

$\therefore 0=u-a_1{T}_1\Rightarrow aT=a_1{T}_1$

and from
$v^2=u^2-2a_1{S}_2\Rightarrow S_2=\frac{u^2}{2a_1}=\frac{1}{2}\frac{a^2{T}^2}{a_1}$

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

Step: 3 Find the average speed of the car.
Formula Used: $v_{av}=\frac{S_1+S_2}{T+T_1}$
$\therefore v_{av}=\frac{S_1+S_2}{T+T_1}$

$=\frac{\frac{1}{2}a{t}^2+\frac{1}{2}aT\times T_1}{T+T_1}$

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