A bus accelerates from 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 seconds then, evaluate. (a) the maximum velocity achieve and (b) the total distance travelled graphically.

A bus accelerates from rest at a constant rate alpha for some time, af

1.	A bus accelerates from 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 seconds then, evaluate. (a) the maximum velocity achieve and (b) the total distance travelled graphically.
Answer» Solution :Let `t_(1)` be the time of acceleration and `t_2` that of DECELERATION of the bus. The total time is `t=t_(1)+t_(2)`. Let `v_("MAX")` be the maximum velocity. As the acceleration and decleratioin are CONSTANTS the velocity time graph is a straight line as shown in the FIGURE with +ve slope for acceleration and -ve slope for decleration. From the graph, the slope of the line OA gives the acceleration `alpha`. `alpha ="slope of the line "OA=(v_("max"))/(t_(1)) rArr t_(1)=(V_("max"))/(alpha)` the slope of AB gives the decleration `beta` `beta="slope of AB"=(v_("max"))/(t_(2)) rArr t_(2)=(v_("max"))/(beta)` `t=t_(1)+t_(2)=(v_("max"))/(alpha)+(v_("max")).(beta)` `t=v_(max) ((alpha+beta)/(alpha beta))` `v_("max")=((alpha beta)/(alpha+beta))t` (b) Displacement =area under the v-t graph =area of `triangle OAB` `=1/2 ("base") ("height") =1/2 t v_("max") =1/2 t ((alpha beta t)/(alpha +beta))=1/2 ((alpha beta t^(2))/(alpha +beta))`