A car starts moving rectilinearly, first with acceleration w = 5.0 ms-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 τ = 25 s. The average velocity during that time is equal to v = 72 kmh-1. How long does the car move uniformly?

Solution:

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 movers uniformly. Then the acceleration/deceleration time is  \( \frac{\tau -\Delta t}{2} \) each. So,

 \( <v>\tau =2\left\{ \frac{1}{2}w\frac{{{(\tau -\Delta \tau )}^{2}}}{4} \right\}+w\frac{(\tau -\Delta t)}{2}\Delta t \)

Or  \( \Delta {{t}^{2}}={{\tau }^{2}}-\frac{4<v>\tau }{w} \)

Hence  \( \Delta t=\tau \sqrt{1-\frac{4<v>}{w\tau }}=15\,\,s \).

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