A car starts moving rectilinearly first with acceleration `alpha=5 m s^(-2)` (the initial velocity is equal to zero), then uniformly, and finally, deceleration at the same rate `alpha` comes to a stop. The time of motion equals `t=25 s`. The average velocity during this time is equal to `
72 kmh^(-1)` How long does the car move uniformly?

Explanation

Time of journey=25 s, Average velocity`=20 m//s`
`"Displacement"="Average velocity"xx"time"`
`=20xx25=500 m`
Since `|a_(1)|=|a_(2)|`, hence `t_(1)=t_(3)=t_(0)(say)`
`t_(2)=25-(t_(1)+t_(3))=25-2t_(0)`
Now sketch the v-t graph (as in the previous problem).
Here, for time `t_(2)`, velocity `v_(m)` is constant.

Slope of the v-t graph` -=` Acceleration
`5=v_(m)/t_(0)impliesv_(m)=5t_(0)`
Area of the v-t graph=Displacement
`1/2v_(m)[25+(25-2t_(0))]=500` (area of trapezium)
`1/2xx5 t_(0)(50-2t_(0))=500`
`t_(0)^(2)-25 t_(0)+100=0`
`(t_(0)-20)(t_(0)-5)=0`
`t_(0)=20` s is not possible `(t_(2)=25-2t_(0)=-ve)`
`t_(0)=5 s`
`t_(2)25-2t_(0)=15 s`