Question

A car starting from rest, accelerates 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 of travel is $t$, find the total distance travelled by it assuming it to travel in a straight line.

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

Answer 1

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

Since acceleration of car $=\alpha =$ constant and
Deceleration of car $=\beta =$ constant

The car starts from rest, hence it's velocity-time graph will be straight line starting from origin $O$ and will contain $+ve$ slope and $-ve$ slope partwise.

Let the car accelerates for time $t_1$ and decelerates for time $t_2$. Then,
$t=t_1+t_2\dots \left(i\right)$
and corresponding velocity-time graph will be as shown in figure:


From the graph:
$\alpha =$ slope of line $OA=\frac{v_{\text{max}}}{t_1}$
or $t_1=\frac{v_{\text{max}}}{\alpha }\dots \left(ii\right)$

$\beta =-$ slope of line $AB=\frac{v_{\text{max}}}{t_2}$
$\Rightarrow t_2=\frac{v_{\text{max}}}{\beta }\dots \left(iii\right)$

From Eqs. $\left(i\right)$, $\left(ii\right)$ and $\left(iii\right)$, we get:
$\frac{v_{\text{max}}}{\alpha }+\frac{v_{\text{max}}}{\beta }=t$ or $v_{\text{max}}\left(\frac{\alpha +\beta }{\alpha \beta }\right)=t$
$\therefore v_{\text{max}}=\frac{\alpha \beta t}{\alpha +\beta }$

Total distance covered
$S=\text{Displacement}=$ Area under $v-t$ graph
($\because$ particle is moving without changing its direction)
$S=\frac{1}{2}\times t\times v_{\text{max}}=\frac{1}{2}\times t\times \frac{\alpha \beta t}{\alpha +\beta }$

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