Question

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

• A. $\frac{\alpha \beta }{2(\alpha +\beta )}{t}^2$
• B. $\frac{\alpha \beta }{4(\alpha +\beta )}{t}^2$
• C. $\frac{4\alpha \beta }{(\alpha +\beta )}{t}^2$
• D. $\frac{2\alpha \beta }{(\alpha +\beta )}{t}^2$

Answer 1

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


During acceleration,
$v_o=0+\alpha t_1=\alpha t_1$

During deceleration,
$0=v_o-\beta t_2\Rightarrow v_o=\beta t_2$

Also, $t_1+t_2=t$

$\Rightarrow \frac{v_o}{\alpha }+\frac{v_o}{\beta }=t$

$\Rightarrow v_o=\frac{\alpha \beta }{\alpha +\beta }t$

Now, total distance travelled,
$=$ Area under speed-time graph

$=\frac{1}{2}\times t\times v_o$

$=\frac{1}{2}\times t\times \frac{\alpha \beta }{\alpha +\beta }t$

$=\frac{\alpha \beta }{2(\alpha +\beta )}{t}^2$