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?

(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 .