I. E. Irodov Solution 1.4

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: 1.4
A point moves rectilinearly in one direction. This figure shows the distance \(S\) traversed by the point as a function of the time \(t\). Using the plot find:
(a) the average velocity of the point during the time of motion.
(b) the maximum velocity.
(c) the time moment \({t_0}\) at which the instantaneous velocity is equal to the mean velocity averaged over the first \({t_0}\) seconds.

Solution: 1.4

(a) The average velocity is distance traveled by time \( = \frac{{200cm}}{{20\sec }} = 10cm/s\)

(b) The maximum velocity is the point where the rate of change of distance that is the slope of the curve is the maximum. In the mid portion of the curve, the point moves \(100cm\) in \(4secs\).
Thus, the slope is \(\frac{{100}}{4}\) = \(25cm/s\). This is the maximum speed achieved.

(c) At any time the instantaneous speed \(\frac{{ds}}{{dt}}\) is the slope of the curve while the average speed \(\frac{s}{t}\) is the tangent of the angle of the line joining the origin to the point.

Let us first consider a time to during the acceleration phase of the point. This is when to is in the interval \(\left( {0,10} \right)\). Throughout this region, the slope of the curve will always be greater than that of the line joining the point to the origin since the curve is convex. This means that throughout the acceleration phase, \(\frac{s}{t}\) will always be less than \(\frac{{ds}}{{dt}}\).

Now lets us consider the region where, the speed is constant. As shown in the picture even in this region the tangent of the line joining the point to the can never catch up the slope of the curve.
However, when we consider a time that is in the decelerating region of the curve, as the instantaneous speed of point decreases, at some time the average speed will catch up with the instantaneous speed. In other words the tangent of angle of line connecting the point to the origin will be same as the slope of the curve. This is shown in the figure beside.

Since exact nature (equation) of the curve of the decelerating part of the curve is not provided (it could be exponentially decaying) it is not possible to mathematically determine the exact value of time at which this happens. However, geometrically one can draw a tangent at every point on the curve and see if it passes through the origin. When this happens it will be the point at which instantaneous speed is same as the average speed.

Here the answer, probably this happens at \(16s\) as suggested answer. Certainly it seems that way but the exact answer cannot be determined mathematically unless the nature of the curve is specified.