A car accelerates from rest at a constant rate α for some time and reaches a maximum velocity. Then it decelerates at a constant rate to come to rest. If the total time lapsed is ‘t’, then the maximum velocity acquired by the car will be     

Given that

Initial velocity u = 0

Uniform acceleration $=\alpha$

Time taken $\text{t}={\text{t}}_{\text{1}}$

Final velocity $\text{v}=\text{u}+a\text{t}$

$\Rightarrow {\text{v}}_{\text{max}}=\alpha {\text{t}}_{\text{1}}$

From that instant it started retarding

So $\text{u}=\alpha {\text{t}}_{\text{1}}$

Acceleration $=-\beta$

Final velocity = v

$\Rightarrow \text{v}=\text{u}+a\text{t}$

$0=\alpha {\text{t}}_{\text{1}}-\beta {\text{t}}_{\text{2}}$

$\therefore {\text{v}}_{\text{max}}=\alpha {\text{t}}_{\text{1}}=\beta {\text{t}}_{\text{2}}$

Let the total time taken $=\text{t}={\text{t}}_{\text{1}}+{\text{t}}_{\text{2}}$

$v_{\max }=\alpha t_1$

$v_{\max }=\beta t_2$

${\text{t}}_{\text{1}}=\frac{v_{\max }}{\alpha },{\text{t}}_{\text{2}}=\frac{v_{\max }}{\beta }$

$\text{t}={\text{t}}_{\text{1}}+{\text{t}}_{\text{2}}$

$\text{t}=v_{\max }\left[\frac{1}{\alpha }+\frac{1}{\beta }\right]$

$\text{t}=v_{\max }\left[\frac{\beta +\alpha }{\alpha \beta }\right]$

$v_{\max }=\left[\frac{\alpha \beta }{\beta +\alpha }\right]\text{t}$

$v_{\max }=\left[\frac{\alpha \beta }{\alpha +\beta }\right]\text{t}=\frac{\alpha \beta }{\alpha +\beta }\text{.t}$