Accelerating and decelerating car

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Problem:
A car starts moving rectilinearly, first with acceleration $w = 5.0 \space \text{m}/\text{s}^2$ (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 $\tau = 25 \text{s}$. The average velocity during that time is equal to $\bar{v} = 72 \text{ km per hour}$. How long does the car move uniformly?



Solution:

Convert $72 \text{ km per hour}$ into $20 \space \text{m}/\text{s}$. Now, everything is in seconds and meters. 

Since the rate of acceleration and deceleration are equal, the car must take the same amount of time and distance in this period.  

Denote $t$ as the time in which the car moves uniformly. Then, the acceleration and deceleration each took $\frac{\tau-t}{2}$ time.

So, we have $$\bar{v}\tau=2\left(\frac{1}{2}w\frac{(\tau-t)^2}{4}\right)+w\frac{(\tau-t)}{2}t$$ $$\implies t^2=\tau^2-\frac{4\bar{v}\tau}{w}$$ $$\implies t=\tau\sqrt{1-\frac{4\bar{v}}{w\tau}}=25\sqrt{1-\frac{4\cdot 20}{5\cdot 25}}$$ $$=\boxed{15\text{ s}}.$$
 

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