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(D) Let the car accelerate at a constant rate $\alpha$ for time $t_1$. The maximum velocity $v$ attained is given by $v = u + \alpha t_1$. Since it st

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$\frac{\alpha \beta t}{\alpha + \beta}$

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(D) Let the car accelerate at a constant rate $\alpha$ for time $t_1$. The maximum velocity $v$ attained is given by $v = u + \alpha t_1$. Since it starts from rest,$u = 0$,so $v = \alpha t_1$. Next,the car decelerates at a constant rate $\beta$ for the remaining time $(t - t_1)$ until it comes to rest. Using the equation $v = u + at$ for the deceleration phase,we have $0 = v - \beta(t - t_1)$. Substituting $v = \alpha t_1$ into the equation,we get $0 = \alpha t_1 - \beta t + \beta t_1$. Rearranging the terms,we get $(\alpha + \beta) t_1 = \beta t$,which gives $t_1 = \frac{\beta}{\alpha + \beta} t$. Substituting $t_1$ back into the expression for $v$,we get $v = \alpha \left( \frac{\beta}{\alpha + \beta} t \right) = \frac{\alpha \beta}{\alpha + \beta} t$.

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

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