A car accelerates from rest at a constant rate $\mathrm{\alpha } \mathrm{for} \mathrm{some} \mathrm{time}, \mathrm{after} \mathrm{which} \mathrm{it} \mathrm{decelerates} \mathrm{at} \mathrm{a} \mathrm{constant} \mathrm{rate} \mathrm{\beta }$ and comes to rest. If the total time elapsed is t, then the maximum velocity acquired by the car is

Let the car accelerate at rate $\mathrm{\alpha } \mathrm{for} \mathrm{time} {\mathrm{t}}_1 \mathrm{then} \mathrm{maximum} \mathrm{velocity} \mathrm{attained}, \mathrm{v} = 0+{\mathrm{\alpha t}}_1 = {\mathrm{\alpha t}}_1$

Now, the car decelerates at a rate $\mathrm{\beta } \mathrm{for} \mathrm{time} (\mathrm{t}-{\mathrm{t}}_1) \mathrm{and} \mathrm{finally} \mathrm{comes} \mathrm{to} \mathrm{rest}. \mathrm{Then},$

$0 = \mathrm{v} - \mathrm{\beta }(\mathrm{t}-{\mathrm{t}}_1) \Rightarrow 0 = {\mathrm{\alpha t}}_1-\mathrm{\beta t}+{\mathrm{\beta t}}_1$

$\Rightarrow {\mathrm{t}}_1 = \frac{\mathrm{\beta }}{\mathrm{\alpha }+\mathrm{\beta }}\mathrm{t}$

$\therefore$v = $\frac{\mathrm{\alpha \beta }}{\mathrm{\alpha }+\mathrm{\beta }}\mathrm{t}$