# A car accelerates from rest at a constant rate α for some time, after which it decelerates at a constant rate β and comes to rest. If the total time elapsed is t then the maximum velocity acquired by the car is

Question : A car accelerates from rest at a constant rate α for some time, after which it decelerates at a constant rate β and comes to rest. If the total time elapsed is t then the maximum velocity acquired by the car is

(a) $\left( \dfrac{\alpha t+\beta }{\alpha \beta }\right) ^{2}t$

(b) $\left( \dfrac{\alpha ^{2}-\beta ^{2}}{\alpha \beta }\right) t$

(c) $\left( \dfrac{\alpha +\beta }{\alpha \beta }\right) t$

(d) $\dfrac{\alpha \beta t}{\alpha +\beta }$

Answer : Option (d)

Solution :

Let the car accelerate for the time t1

Then , the maximum velocity reached is

V = 0 + α (t1)

Now the car decelerates and then finally comes to rest

0 = v – β ( t2 )

Therefore,

$t_{1}=\dfrac{v}{\alpha },t_{2}=\dfrac{v}{\beta }$

$t=t_{1}+t_{2}\Rightarrow t=\dfrac{v}{\alpha }+\dfrac{v}{\beta }\Rightarrow t=v\left( \dfrac{1}{\alpha }+\dfrac{1}{\beta }\right)$

\begin{aligned}t=V\left( \dfrac{\alpha +\beta }{\alpha \beta }\right) \Rightarrow \ V=\dfrac{\alpha \beta ^{t}}{\alpha +\beta }\end{aligned}
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