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in Physics by (71.3k points)

Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 50 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?

1 Answer

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(a) From end A to end B

Distance covered by Joseph while jogging from A to B = 300 m

Time taken to cover that distance = 2 min 50 seconds = 170 s

Total distance covered = 300 m

Total time taken = 170 s

Displacement = shortest distance between A and B = 300 m

Time interval = 170 s

Average speed = 300/170 = 1.765 m/s

Displacement = shortest distance between A and B = 300 m

Time interval = 170 s

Average velocity = 300/170 = 1.765 m/s

The average speed and average velocity of Joseph from A to B are the same and equal to 1.765 m/s.

(b) From end A to end C

Total distance covered = Distance from A to B + Distance from B to C

= 300 + 100 = 400 m

Total time taken = Time taken to travel from A to B + Time taken to travel from B to C 

= 170 + 60 = 230 s

Average speed = 400/230 = 1.739 m/s

Displacement from A to C = AC = AB − BC 

= 300 − 100 = 200 m

Time interval = time taken to travel from A to B + time taken to travel from B to C

= 170 + 60 = 230 s

Average velocity = 200/230 = 0.87 m/s

The average speed of Joseph from A to C is 1.739 m/s and his average velocity is 0.87 m/s.

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