A car acceleration form 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 elapsed is t evaluate (a) the maximum velocity attained and (b) the total distance travelled.

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.")`