Question

A car accelerates from rest at a constant rate $\alpha$ for some time after which it decelerates at a constant rate $\beta$ to come to rest. If the total time elapsed is t then, the distance travelled by the car is

• A. $\frac{1}{2}(\frac{\alpha \beta }{\alpha +\beta })t^2$
• B. $\frac{1}{2}(\frac{\alpha +\beta }{\alpha \beta })t^2$
• C. $\frac{1}{2}(\frac{{\alpha }^2+{\beta }^2}{\alpha \beta })t^2$
• D. $\frac{1}{2}(\frac{{\alpha }^2-{\beta }^2}{\alpha \beta })t^2$

Answer 1

The correct option is A $\frac{1}{2}(\frac{\alpha \beta }{\alpha +\beta })t^2$

Lets consider, $v=$ maximum velocity,

When A car accelerates from rest at a constant rate $\alpha$

From 1st equation of motion,

$v=u+at$

$v=0+\alpha t_1$

$v=\alpha t_1$

$t_1=\frac{v}{\alpha }$

After car decelerate at the rate $\beta$,

From 1st equation of motion,

$v=u+at$

$0=v-\beta t_2$

$v=\beta t_2$

$t_2=\frac{v}{\beta }$

Total time elapsed,

$t=t_1+t_2$

$t=\frac{v}{\alpha }+\frac{v}{\beta }$

$t=v\left(\frac{\alpha +\beta }{\alpha \beta }\right)$

$v=\left(\frac{\alpha \beta }{\alpha +\beta }\right)t$

From the above figure,

The total distance covered is equal to the area under the velocity-time graph

$S=\frac{1}{2}\times v\times t$

$s=\frac{1}{2}\left(\frac{\alpha \beta }{\alpha +\beta }\right)t^2$

The correct option is A.