1.

Abus accelerates from rest at a constant rate alpha for some time,after which it decerates at a constant rate beta to come to rest.If the total time elapes is t seconds then ,evaluate. (a)the maximum velocity achieved and (b)the total distance travelled graphically.

Answer»

Solution :Let `t_(1)` be the time of acceleration and `t_(2)` that of decelration 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 CONSTANTS 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` =slope of the line `OA=(v_(max))/(t_(1))impliest_(1)=(v_(max))/(alpha)`
The slope of AB gives the deceleration `beta`
`thereforebeta=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)/(alphabeta))`
`thereforev_(max)=((alphabeta)/(alpha+beta))t`
Displacement =area under the v-t graph `=(1)/(2)(base)(height)=(1)/(2)t v_(max)=(1)/(2)t((alphabetat)/(alpha+beta))=(1)/(2)((alphabetat^(2))/(alpha+beta))`


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