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.

Explanation

(i) For the time `t_(1)`,
v=u+at
`=0+xt_(1)`
`rArr t_(1)=(v)/(x)`
here v will be the maximum velocity attained by the train because after that, it will start to decelerate
For the time `t_(2)=(t-_(1))`
or `t-t_(1)=(v)/(y)`
Adding i and ii we get,
`t_(1)+t-t_(1)=(v)/(x)+(v)/(y)`
`rArr v((1)/(x)+(1)/(y))=v((x+y)/(xy))`
or `v=(xyt)/(x+y)`
This is the required expression for maximum velocity (ii) Distance travelled by the train in time `t_(1)` is
`s_(1)=ut+(1)/(2)at^(2)`
`=0+(1)/(2)st_(1)^(2)=(1)/(2)x(v^(2))/(x^(2))=(v^(2))/(2x)`
`=0+(1)/(x)st_(1)^(2)=(1)/(2)x(v^(2))/(x^(2))=(v^(2))/(2x)`
`=(x^(2)y^(2)t^(2))/(2x(x+y)^(2))`
`=(xy^(2)t^(2))/(2(x+y)^(2))`
Distance travelled by the train in time `t_(2)=(t-t_(1))` is
`s_(2)=ut+(1)/(2)at^(2)`
`=v(t-t_(1))-(1)/(2)y(t-t_(1))^(2)`
`=v(t-t_(1))-(1)/(2)y(t-t_(1))^(2)`
`=v(v)/(y)-(1)/(2)y(v^(2))/(y^(2))=(v^(2))/(2y)=(x^(2)yt^(2))/(2(x+y)^(2))`
`therefore` Total distance travelled by the train
`s=s_(1)+s_(2)=(xy^(2)t^(2))/(2(x+y)^(2))+(x^(2)yt^(2))/(2(x+y)^(2))`
`=(xyt^(2))/(2(x+y)^(2))(y+x)`
`=(xyt^(2))/(2(x+y))`