A car accelerates from rest at a constant rate for some time after which it decelerates at a constant rate `beta` to come to rest. If the total time elapsed is t, the maximum velocity acquired by the car is given by :

Options

A. `(alpha beta)/(t(alpha + beta))`
B. `(alpha + beta) /(alpha beta) t `
C. `(alpha beta)/(alpha + beta) t `
D. `alpha beta)/((alpha + beta))`

Explanation

If the car accelerates for time `t_(1)` and decelerates for time `t_(2)` , then according to given problems
` t = t_(1) + t_(2)`
If ` v_("max")` is the maximum valocity of the car , then from v/t curve , we have
` alpha = (V_("max"))/(t_(1)) , beta = (v_("max"))/(t_(2))`
[as slope of v/t curve gives acceleration.]
so ` ((1)/(alpha ) + (1)/(beta)) = (t_(1) + t_(2))/(v_("max")) rArr V_("max") = (alpha beta) /((alpha + beta)) t `

` [ because t = t_(1) + t_(2)]`
Hence correct answer is (C) .