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Question

Physics Question on Motion in a straight line

A car accelerates from rest at a constant rate α\alpha for some time after which it decelerates at a constant rate β\beta and comes to rest. If total time elapsed is tt, then maximum velocity acquired by car will be

A

(α2+β2αβ)t\left(\frac{\alpha^{2}+\beta^{2}}{\alpha \beta}\right) t

B

(α2β2αβ)t\left(\frac{\alpha^{2}-\beta^{2}}{\alpha \beta}\right) t

C

(α+βαβ)t\left(\frac{\alpha+\beta}{\alpha \beta}\right) t

D

(αβα+β)t\left(\frac{\alpha \beta}{\alpha+\beta}\right) t

Answer

(αβα+β)t\left(\frac{\alpha \beta}{\alpha+\beta}\right) t

Explanation

Solution

Let maximum velocity =v= v Now, v=0+αt1v = 0 + \alpha t _{1}
Similarly, 0=vβt20= v - \beta t _{ 2 }
From the above equations we get,
t1=vα&t2=vβt _{1}=\frac{ v }{ \alpha } \quad \& \quad t _{2}=\frac{ v }{ \beta }
t1+t2=t=vα+vβt _{1}+ t _{ 2 }= t =\frac{ v }{ \alpha }+\frac{ v }{ \beta }
v=αβα+βt\Rightarrow v =\frac{\alpha \beta}{ \alpha + \beta } t