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, the total distance travelled by the car is given by

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

Answer 1

The correct option is A

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

Let $t_1$ be the time during which the car accelerates at a rate $\alpha$. The velocity at the end of time $t_1$ will be $v=u+\alpha t_1=0+\alpha t_1=\alpha t_1(\because u=0)$
The time during which the car decelerates is $t_2=t-t_1$. For this part, the initial velocity is $\alpha t_1$ and the final velocity is zero and the acceleration is $-\beta$.
$\therefore \alpha t_1=\beta (t-t_1)$
$\Rightarrow t_1=\frac{\beta t}{\alpha +\beta }$
The distance travelled in time $t_1$ is
$s_1=u{t}_1+\frac{1}{2}\alpha t_1^2=\frac{1}{2}\alpha t_1^2=\frac{1}{2}\alpha {\left(\frac{\beta }{\alpha +\beta }\right)}^2{t}^2$
Also, the distance travelled in time $t_2$ is given by
$0^2={\left(\alpha t_1\right)}^2-2\beta s_2\Rightarrow 2\beta s_2={\alpha }^2{t_1}^2={\left(\frac{\alpha \beta }{\alpha +\beta }\right)}^2{t}^2\Rightarrow s_2=\frac{1}{2}\beta {\left(\frac{\alpha }{\alpha +\beta }\right)}^2{t}^2$
$\therefore s_1+s_2=\frac{1}{2}\alpha {\left(\frac{\beta }{\alpha +\beta }\right)}^2{t}^2+\frac{1}{2}\beta {\left(\frac{\alpha }{\alpha +\beta }\right)}^2{t}^2=\frac{1}{2}\frac{\alpha \beta }{\alpha +\beta }{t}^2$
Hence, the correct choice is (a).