Question

Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 50 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?

Answer 1

Total distance covered from A to B $=300m$
Total time taken $=2\times 60s+50s$ $=170s$

Therefore, average speed from A to B $=\frac{TotalDistance}{TotalTime}$ $=\frac{300}{170}$ $=1.76m{s}^{-1}$
Average velocity from A to B $=\frac{DisplacementAB}{Time}=\frac{300}{170}$ $=1.76m{s}^{-1}$
Total distance covered from A to C $=AB+BC$ $=300+100m=400m$
Total time taken from A to C = $=170s+60s$ $=230s$
Therefore, average speed from A to C $=\frac{TotalDistance}{TotalTime}$ $=\frac{400}{230}m{s}^{-1}$ $=1.74m{s}^{-1}$
Displacement from A to C = AC $=200m$
Time taken for displacement from A to C = 230 s
Therefore, average velocity from A to C $=\frac{Displacement}{Time}$ $=\frac{200}{230}m{s}^{-1}$ $=0.87m{s}^{-1}$