InterviewSolution
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InterviewSolution
| 1. |
Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minute 30 second and then turns around and jogs 100 m back to point 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? |
Answer» Solution : (a) From A to B, distance COVERED = 300 m = displacement, time taken = 2 min 30 SEC `= 2 xx 60 + 30 = 150 sec ` Average speed `= ("TOTAL distance covered")/("Total time taken")` `= (300 m )/( 150 s ) = 2 ms ^(-1)` Average velocity `= ("Displacement ")/("Time taken")` `= (300 m)/(150)s ) = 2 ms ^(-1)` (b) From A to B to C, distance covered `= (300 + 100) m = 400 m` displacemenet `= AB -BC = (300 -100) = 200 m` time taken `= 2 min 30 sec + 1 min = (150 + 60) s = 210 s` Average speed `= ("Total distance covered")/("Total time taken")` `= (400)/(210) ~~ 1.90 ms ^(-1)` Average velocity `= ("Displacement")/("Time taken") = (200 )/(210) ~~ 0.95 ms ^(-1)` (a) Average speed `=1.90 ms ^(-1)` Average velocity `= 0.95 m s ^(-1)` |
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