Exercise 4.5
Under what condition(s) is the (i) magnitude of average velocity of an object equal to its average speed? (ii) magnitude of average velocity of an object zero while its average speed is not zero?
Not cross-checked
NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.
(i) They are equal when the object moves in a straight line in one direction only, never turning back — then the total distance travelled equals the magnitude of displacement, so dividing either by the same time interval gives the same number.
(ii) The magnitude of average velocity is zero while the average speed is not, when the object returns to its starting position by the end of the time interval — the displacement is \(\displaystyle 0\ \text{m} \) but the path actually covered is not zero.
Worked example for (ii): Sarang swims $\displaystyle 25$ m and back in $\displaystyle 50$ s (Example $\displaystyle 4.2$), so \(\displaystyle v_{av} = \frac{0\ \text{m}}{50\ \text{s}} = 0\ \text{m s}^{-1} \) while \(\displaystyle \text{average speed} = \frac{50\ \text{m}}{50\ \text{s}} = 1\ \text{m s}^{-1} \).
One complete revolution on a circular path is another such case: distance \(\displaystyle = 2\pi R \), displacement \(\displaystyle = 0 \).