Question

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 1

(i) 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 figure.

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)`
(ii) Total distance = displacement = area under v - t graph
`=(1)/(2)xxtxxv_(max)=(1)/(2)xxtxx(alpha beta t)/(alpha+beta)`
or Distance `=(1)/(2)((alpha beta t^(2))/(alpha+beta))`