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(A) Let $t_1$ be the time taken to accelerate at rate $\alpha$ to reach maximum velocity $v_{\max}$,and $t_2$ be the time taken to retard at rate $\be

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $t_1$ be the time taken to accelerate at rate $\alpha$ to reach maximum velocity $v_{\max}$,and $t_2$ be the time taken to retard at rate $\beta$ to come to rest.From the velocity-time graph,the slope of the acceleration phase is $\alpha = \frac{v_{\max}}{t_1}$,which gives $t_1 = \frac{v_{\max}}{\alpha}$.The slope of the retardation phase is $\beta = \frac{v_{\max}}{t_2}$,which gives $t_2 = \frac{v_{\max}}{\beta}$.The total time is $t = t_1 + t_2$.Substituting the expressions for $t_1$ and $t_2$: $t = \frac{v_{\max}}{\alpha} + \frac{v_{\max}}{\beta}$.$t = v_{\max} \left( \frac{\alpha + \beta}{\alpha \beta} \right)$.Solving for $v_{\max}$,we get $v_{\max} = \frac{\alpha \beta t}{\alpha + \beta}$.

$A$ car starts from rest and travels with uniform acceleration $\alpha$ for some time and then with uniform retardation $\beta$ and comes to rest. If the total travel time of the car is $t$,the maximum velocity attained by it is given by:

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