Question

A car starts moving rectilinearly, first with acceleration $w=5.0m/s^2$ (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 $\tau =25s$. The average velocity during that time is equal to $⟨v⟩=72km/h$. The car moves uniformly for (10+x) seconds. The value of x is:

Answer 1

As the car starts from rest and finally comes to a stop, and the rate of acceleration and deceleration are equal, the distances as well as the times taken are same in these phases of motion.
Let $\Delta t$ be the time for which the car moves uniformly. Then the acceleration/deceleration time is $\frac{\tau -\Delta t}{2}$ each. So total distance is calculated as
$s=2(ut+\frac{1}{2}a{t}^2)+v_{max}t$
$v_{max}$ is calculated using $v=u+at=0+w\frac{\tau -\Delta t}{2}$
Thus we get
$⟨v⟩\tau =2\left\{\frac{1}{2}w\frac{{(\tau -\Delta t)}^2}{4}\right\}+w\frac{(\tau -\Delta t)}{2}\Delta t$
$=w\frac{\tau -\Delta t}{2}(\frac{\tau -\Delta t}{2}+\Delta t)$
$=w\frac{\tau -\Delta t}{2}\left(\frac{\tau +\Delta t}{2}\right)$
Solving it we get
$\Delta t^2={\tau }^2-\frac{4⟨v⟩\tau }{w}$
Hence $\Delta t=\tau \sqrt{1-\frac{4⟨v⟩}{w\tau }}=15s$.