Question

A car starts moving rectilinearly, first with acceleration w = 5.0 m/s2 (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate w, comes to a stop. The total time of motion equals τ = 25 s. The average velocity during that time is equal to <v> = 72 km per hour. How long does the car move uniformly?

Answer 1

Solution

According to given question car starts from rest and then comes to a stop by same acceleration and deacceleration and also time taken and distances travelled are same.

Now consider dt is time when car moves uniformly so acceleration/deacceleration time is given by as $\frac{(t-dt) }{2}$ .

So,

$<v>\tau=2\frac{w(\tau-dt) ^2}{8}+\frac{w(\tau-dt).dt}{2}$

$(dt) ^2=\tau^2-\frac{4<v>\tau}{w}$

$(dt) =\tau\sqrt{1-\frac{4<v>}{w\tau}}$

Now putting value of <v>=72km/h,

$\tau=25s,$ $w=5m/s^2$

Therefore car moves uniformly by time

$dt=15sec.$