Joseph jogs from one end A to another end B of a straight 300 m road in 2 minutes and 30 sec
Joseph jogs from one end A to another end B of a straight 300 m road in 2 minutes and 30 sec 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
(b) from A to C?
1 Answers
(a) For motion from A to B:
Distance covered = 300 m
Displacement = 300 m.
Time taken = 150 sec.
We know that, Average speed = Total distance covered ÷ Total time taken
= 300 m ÷ 150 sec = 2 ms-1
Average velocity = Net displacement ÷ time taken
= 300 m ÷ 150 sec = 2 ms-1
(b) For motion from A to C:
Distance covered = 300 + 100 = 400 m.
Displacement = AB - CB = 300 - 100 = 200 m.
Time taken = 2.5 min + 1 min = 3.5 min = 210 sec.
Therefore, Average speed = Total distance covered ÷ Total time taken
= 400 ÷ 210 = 1.90 ms-1 .
Average velocity = Net displacement ÷ time taken
= 200 m ÷ 210 sec = 0.952ms-1 .