- Let the car accelerate at a rate $\alpha$ for time t1 and attain a maximum velocity v. Then the car decelerates at a constant rate $\beta$ for remaining time (t - t1) and again comes to rest. Then from adjoining figure, it is clear that
$\text{v}=\alpha\text{t}_1=\beta(\text{t}-\text{t}_1)$
$\therefore \text{t}_1=\frac{\text{v}}{\alpha}$
$(\text{t}-\text{t}_1)=\frac{\text{v}}{\beta}$
Adding these two, we have
$\text{t}=\frac{\text{v}}{\alpha}+\frac{\text{v}}{\beta}=\text{v}\Big(\frac{\alpha+\beta}{\alpha\beta}\Big)$
$\Rightarrow \text{v}=\frac{\alpha\beta\text{t}}{(\alpha+\beta)}$
- Total distance travelled by the car s = area OAB $=\frac{1}{2}(\text{t})\times(\text{v})$
$\therefore \text{s}=\frac{1}{2}\text{t}\times\frac{\alpha\beta\text{t}}{(\alpha+\beta)}$
$=\frac{1}{2}\frac{\alpha\beta}{(\alpha+\beta)}\text{t}^2$
