Question

5. George walks to a friend's house. He walks 750 meters North, then realizes he walked too far.

Answer 1

Answer:

Average speed: approximately $76.9\; {\rm m\cdot s^{-1}}$.

Average velocity: approximately $38.5\; {\rm m \cdot s^{-1}}$ (to the north.)

Explanation:

Consider an object that travelled along a certain path. Distance travelled would be equal to the length of the entire path.

In contrast, the magnitude of displacement is equal to distance between where the object started and where it stopped.

In this question, the path George took required him to travel $750\; {\rm m} + 250\; {\rm m} = 1000\; {\rm m}$ in total. Hence, the distance George travelled would be $1000\; {\rm m}$. However, since George stopped at a point $(750\; {\rm m} - 250\; {\rm m}) = 500\; {\rm m}$ to the north of where he started, his displacement would be only $500\; {\rm m}$ to the north.

Divide total distance by total time to find the average speed.

Divide total displacement by total time to find average velocity.

The total time of travel in this question is $13\; {\rm s}$.. Therefore:

$\begin{aligned}\text{average speed} &= \frac{\text{total distance}}{\text{total time}} \\ &= \frac{1000\; {\rm m}}{13\; {\rm s}} \\ &\approx 76.9\; {\rm m\cdot s^{-1}}\end{aligned}$.

$\begin{aligned}\text{average velocity} &= \frac{\text{total displacement}}{\text{total time}} \\ &= \frac{500\; {\rm m}}{13\; {\rm s}} && \genfrac{}{}{0px}{}{(\text{to the north})}{}\\ &\approx 38.5\; {\rm m\cdot s^{-1}} && (\text{to the north})\end{aligned}$.