Joseph jogs from one end A to the other end B of a straight 300m road in 2min.30 s and then turnsaround and jogs 100m back to point C in another 1min. What are Joseph’s average speeds and velocitiesin jogging from A to B and from A to C?
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Explanation:
ANSWER
From pont A to B
Distance covered=300m
Displacement=300m
Time taken=2 min 30 sec=(2×60)+30=150 sec
Average speed=
Timetaken
Distance covered
=
150
300
=2 m/sec
Average velocity=
Timetaken
Displacement
=
150
300
=2 m/sec
From point A to C
Distance covered=300+100=400m
Displacement=300−100=200m
Time taken=3 min 30 sec=(3×60)+30=210 sec
Average speed=
Timetaken
Distance covered
=
210
400
=1.90 m/sec
Average velocity=
Timetaken
Displacement
=
210
200
=0.95 m/sec
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