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Question 2
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?

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Solution

Total distance covered from A to B =300 m
Total time taken =2×60 s+50 s =170 s

Therefore, average speed from A to B =Total DistanceTotal Time =300170 =1.76 ms1
Average velocity from A to B =Displacement ABTime=300170 =1.76 ms1
Total distance covered from A to C =AB+BC =300+100 m=400 m
Total time taken from A to C = =170 s+60 s =230s
Therefore, average speed from A to C =Total DistanceTotal Time =400230 ms1 =1.74 ms1
Displacement from A to C = AC =200m
Time taken for displacement from A to C = 230 s
Therefore, average velocity from A to C =DisplacementTime =200230 ms1 =0.87 ms1

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