A boy is running on a straight road.He runs 500m towards north in 2minutes 10seconds and then turns back and runs 200m in 1.00 minute.Calculate his average speed and magnitude of average velocity during first 2minutes 10seconds and his average speed and magnitude of average velocity during the whole journey.
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Answered by
294
Ahoy !
Let the boy start from a point A and B point is 500 m away from it .
And a C point which is 200 m away from point B .
Now
Time taken to reach at point B.
T= 130s
Distance = 500m
avg speed = 500/ 130
= 3.7m/s
displacement = 500m
v.avg = 500 / 130
= 50 / 13
= 3.7 m/s
Total joutney's avg speed
= (500/ 130 + 200 / 60 ) / 2
= ( 3.7 + 3.3 ) / 2
= 7 / 2
= 3. 5 m/s
as we know the AC = 300m
And time taken from reaching A to C is
= 130 - 60 sec
= 70 sec
Total avg velocity
= 300 / 70
= 4.2 m/s
May it helps you :)
Let the boy start from a point A and B point is 500 m away from it .
And a C point which is 200 m away from point B .
Now
Time taken to reach at point B.
T= 130s
Distance = 500m
avg speed = 500/ 130
= 3.7m/s
displacement = 500m
v.avg = 500 / 130
= 50 / 13
= 3.7 m/s
Total joutney's avg speed
= (500/ 130 + 200 / 60 ) / 2
= ( 3.7 + 3.3 ) / 2
= 7 / 2
= 3. 5 m/s
as we know the AC = 300m
And time taken from reaching A to C is
= 130 - 60 sec
= 70 sec
Total avg velocity
= 300 / 70
= 4.2 m/s
May it helps you :)
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35
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