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Joseph jogs from one end A to other end B of a straight 300m 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?

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Answer Text

Answer :
(a) average speed=average velocity `=2.00 m s^(-1)` <br> (b) average speed `=1.90 m s^(-1)`; average velocity `=0.958 m s^(-1)`
Solution :
Total Distance covered from AB = 300m <br> Total time taken = `2xx60+30` s <br> = 150 s <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_SCI_IX_C08_E01_029_S01.png" width="80%"> <br> Therefore, Average Speed from AB = Total Distance / Total Time <br> = 300/150 m `s^(-1)` <br> = 2 m `s^(-1)` <br> Therefore, Velocity from AB = Displacement AB / Time = 300/ 150 m `s^(-1)` <br> = 2 m `s^(-1)` <br> Total Distance covered from AC = AB + BC <br> = 300 + 200 m <br> Total time taken from A to C = Time taken for AB + Time taken for BC <br> = (`2xx60+30`)+60 s <br> = 210 s <br> Therefore, Average Speed from AC = Total Distance /Total Time <br> = 400/210 m `s^(-1)` <br> = 1.904 m `s^(-1)` <br> Displacement (S) from A to C = AB - BC <br> = 300-100 m <br> = 200 m <br> Time (t) taken for displacement from AC = 210s <br> Therefore, Velocity from AC = Displacement (s) / Time (t) <br> = 200/210 m `s^(-1)` <br> = 0.952 m `s^(-1)`

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