Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 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?
(a) From A to B.
Time for A to B = 2 min 30 s
= 2 x 60 + 30 = 150 s
Average speed = \(\frac{total\;distance}{time\;interval}\)
\(\frac{300}{150}\) = 2 m s-1
Average speed = \(\frac{displacement}{time\;interval}\)
\(\frac{300}{150}\) = 2 m s-1
(b) From A to C.
Time taken = A to B + B to C, 150 + 60 = 210 s
Total distance = 300 + 100 = 400 m
∴ Average speed = \(\frac{displacement}{time\;interval}\)
\(\frac{400}{210}\) = 1.9 m s-1
∴ Average velocity = \(\frac{displacement}{time\;interval}\)
\(\frac{200}{210}\) = 0.95 m s-1