A car accelerates from rest at a constant rate `alpha` for some time afrer which it dccelerates at a constant rate of `beta` to come to rest , If the total time elapsed is `t` second, then calculat
a. the maximum velocity attained by the car, and
b. the total displacement travelled by the car in terms of `alpha , beta` and `t`.

Explanation

(a) Let the car accelerates for time `t_(1)` and decelerates for time `t_(2)` then `t=t_(1)+t_(2)` …(i)
and corresponding velocity-time graph will be as shown in figure
From the graph `alpha=` slope of line `OA=v_("max")/t_(1)rArr t_(1)=v_("max")/alpha`
and `beta=-` slope of line `" " AB=v_("max")/t_(2)rArr t_(2)=v_("max")/beta`
`rArr v_("max")/alpha+v_("max")/beta=trArr v_("max")((alpha+beta)/(alpha beta))=t=trArr v_("max")=(alpha beta t)/(alpha +beta)`

(b) Total distance = area under v-t graph `=1/2txxv_("max")=1/2 t (alphabeta t)/(alpha+beta)=1/2 ((alphabeta t^(2))/(alpha+beta))`
Note : This problem can also be solved by using equations of motion `(v=u+at, "etc.")`