I. E. Irodov Solution 1.3

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Problem: 1.3

A car starts moving rectilinearly, first with acceleration $\omega = 5.0m{s^{ - 2}}$ (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate $\omega$, comes to a stop. The total times of motion equals $\tau = 25s$. The average velocity during that time is equal to $\left\langle v \right\rangle = 72km/h$. How long does the car move uniformly?

Solution: 1.3



Mean velocity is total distance by total time. The $v - t$ graph is shown in the figure above.
Suppose that time of constant acceleration was $t$. During the initial acceleration phase thus the car will travel $\frac{1}{2}\omega {t^2}$ and its final speed will be $\omega t$. This can also be calculated using the $v - t$ diagram as the area under the line $AB$ ($\Delta ABE$).
The car will take the same time $t$ to come to complete rest and during this deceleration phase it will travel $\left( {\omega t} \right)t - \frac{1}{2}\omega {t^2} = \frac{1}{2}\omega {t^2}$.
This can be calculated in the $v - t$ diagram as the area of the triangle $\Delta CFD$.
During the uniform motion phase the car travels at a speed $\omega t$ and travels for the remaining time of $\left( {\tau - 2t} \right)$.
Thus, during the uniform motion phase it travels a distance of $\omega t\left( {\tau - 2t} \right)$.
This can be calculated as the area of the rectangle $EBCF$.
The total distance traveled is thus given by,
$\frac{1}{2}\omega {t^2} + \omega t\left( {\tau - 2t} \right) + \frac{1}{2}\omega {t^2} = \omega t\tau - \omega {t^2}$

The average velocity during the entire time is thus given by,
$\frac{{\omega t\tau - \omega {t^2}}}{\tau } = \left\langle v \right\rangle$
or, $\omega {t^2} - \omega t\tau + \left\langle v \right\rangle \tau = 0$
The above equation is a quadratic equation which has two possible solution,
$t = \frac{{\omega \tau \pm \sqrt {{\omega ^2}{\tau ^2} - 4\omega \left\langle v \right\rangle \tau } }}{{2\omega }}$

Clearly we choose the negative sign since $t$ cannot exceed the total time.
The uniform interval is thus given by,
$\tau - 2t = \tau \sqrt {1 - \frac{{4\left\langle v \right\rangle }}{{\omega \tau }}}$