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?

Answer

`Deltat=tausqrt(1-4 < < v > > //omegatau)=15s`.

Explanation

As the car starts from rest and finally comes to a stop, and the rate of acceleration and decleration are equal, the distance as well as the time taken are same in these phases of motion.
Let `Deltat` be the time for which the car moves uniformly. Then the acceleration/declaration time is `(tau-Deltat)/(2)` each. So,
`lt v gt tau=2{1/2w((tau-Deltat)^2)/(4)}+w((tau-Deltat))/(2)Deltat`
or `Deltat^2-tau^2-(4lt v gttau)/(w)`
Hence `Deltat=tausqrt(1-(4lt v gt)/(wtau))=15s`.