I. E. Irodov Solution 1.3

Solution to I. E. Irodov in General Physics and H. C. Verma in Concept of Physics is like a bible for student who are appearing for IIT-JEE, JEE/Main and JEE/Advance, UG NEET, AIIMS or any other Engineering and Medical entrance examination. All the questions in these books are of high level, which requires all basics to applied concept of physics. We here makes your task very easy, we have presented complete solution with detailed explanation step by step. The solution of these books teaches students also teachers in a suitable manner and then tests you with some tricky questions. To answer these questions you need to have thorough understanding of the concepts and this is where most students falter.

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,
\(\tau - 2t = \tau \sqrt {1 - \frac{{4\left\langle v \right\rangle }}{{\omega \tau }}} \)