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}