Question

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’* average speeds and velocities in jogging

(a) from A to B

(b) from A to C?

Answer 1

(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 .