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 correspoinding velocity-time graph will be as shown in. fig.
From the graph `alpha=` slope of line `OA=(v_("max"))/(t_(1))` or `t_(1)=(v_("max"))/(alpha)` .....(ii)
and `beta=0` slope of line `AB=(v_("max"))/(t_(2))` or `t_(2)=(v_("max"))/(beta)` ....(iii)
from Eqs. (i),(ii) and (iii) `(v_("max"))/(alpha)+(v_("max"))/(beta)=t` or `v_("max")((alpha+beta)/(alpha beta))=t` or `v_("max")=(alpha beta t)/(alpha+beta)`
(b) Total distance = area under v-t graph `=(1)/(2)xx t xx v_("max")=(1)/(2)xxtxx(alpha beta t)/(alpha+beta)`
Distance `=(1)/(2)((alphabeta)/(alpha+beta))t^(2)`