Question

A car accelerates from rest at a constant rate $\alpha$ for some time after which it decelerates at a constant rate $\beta$ to come to rest. If the total time of journey is $t$, then the maximum velocity acquired by the car is given by

• A. $\left(\frac{\alpha +\beta }{\alpha \beta }\right)t$
• B. $\frac{\alpha \beta }{\alpha +\beta }t$
• C. $\frac{{\alpha }^2-{\beta }^2}{\alpha \beta }t$
• D. $\frac{\alpha \beta }{\alpha -\beta }t$

Answer 1

The correct option is B $\frac{\alpha \beta }{\alpha +\beta }t$

Let the car accelerates for time $t_1$ and decelerates for time $t_2$. Then $t_1+t_2=t$
For the first journey
$V_{\text{max}}=0+a{t}_1$
$t_1=\frac{V_{\text{max}}}{\alpha }$
For the second journey
$0=V_{\text{max}}-\beta t_2$
$t_2=\frac{V_{\text{max}}}{\beta }$
Then the total time taken is
$\frac{V_{\text{max}}}{\alpha }+\frac{V_{\text{max}}}{\beta }=t$
$V_{\text{max}}=\frac{\alpha \beta t}{\alpha +\beta }$
Therefore option b is correct.