2. Joseph jogs from one end to the other end B of a straigh 300 m road in 2 minutes 30 seconds and then turns aroun and jogs 100 m back to point C in another 1 minute. What ar Joseph's average speeds and velocities in jogging (a) from Ain B and (b) from A to C?
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Answer:
Total distance covered from AB = 300 m
Total time taken = 2 × 60 + 30 s = 150 s
Therefore, average speed from AB
Total Time= 150sec
Total Distance = 300m
Speed= 300/150 = 2m/s
Therefore, velocity from AB
Time
Displacement AB =
150
300 = ms–1
= 2 m s–1
Total distance covered from AC
= AB + BC
= 300 + 200 m
Total time taken from A to C
= Time taken for AB + Time taken for BC
= (2 × 60 + 30) + 60 s = 210 s
Therefore, average speed from AC
Total Time
Total Distance =
210
400 = m s–1
= 1.904 m s–1
Displacement (s) from A to C
= AB — BC
= 300 – 100 m = 200 m
Time (t) taken for displacement from AC = 210 s
Therefore, velocity from AC
( )
( )
Time t
Displacement s = 210
200 = m s–1=0.952 m s–1
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