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 achieved 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 achieved and (b) the total distance travelled graphically.
Answer» Solution :(a) Let `t_(1)` bethe 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 deceleration are constant the velocity time GRAPH is a straight line as shown in the figure with `+ve` slope for acceleration and `-ve` slope for deceleration. From the graph, the slope of the line OA gives the acceleration `ALPHA`. `therefore alpha` = slpe of the line `OA=(v_(max))/(t_(1))impliest_(1)=(v_(max))/(alpha)` the slope of AB gives the deceleration `beta` `therefore beta` = slope of `AB=(v_(max))/(t_(2))impliest_(2)=(v_(max))/(beta)` `t=t_(1)+t_(2)=(v_(max))/(alpha)+(v_(max))/(beta)` `t=v_(max)((alpha+beta)/(alphabeta))` `therefore v_(max)=((alphabeta)/(alpha+beta))t` (b) DISPLACEMENT = area under the v-t graph = area of `DeltaOAB` `=(1)/(2)("base")("height")=(1)/(2)tv_("max")` `=(1)/(2)t((alphabetat)/(alpha+beta))=(1)/(2)((alphabetat^(2))/(alpha+beta))`