1.

A car 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 evalute (a) the maximum velocity reached and (b) the total distance travelled.

Answer» Correct Answer - A::B::C::D
(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 fig.
`(DCP_V01_CH6_S01_024_S01.png" width="80%">
From the graph, `alpha`=slope of line `OA=v_(max)/t_1` or `t_1=v_(max)/alpha....(ii)`
and `beta` = -slope of line `AB=v_(max)/t_2`
or `t_2=v_(max)/beta...(iii)`
From Eqs. (i),(ii) and (iii), we get
`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 =Total displacement=area under v-t graph
`=1/2xxtxxv_(max)`
`=1/2xxtxx(alpha beta t)/(alpha+beta)`
or Distance = ` 1/2((alpha beta t)/(alpha+beta))`


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