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’s 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 .