From t = 0 to t =5 min, a man stands still, and from t = 5 min to t = 10 min, he walks briskly in a straight line at a constant speed of 2.20m/s. What is avg. velocity and avg. acceleration in the time interval of 2min to 8min.
a) What are his averages velocity and average acceleration from 3min to 8 min
b) 3 min to 9min
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1 response so far ↓
1 Somu // Sep 21, 2008
Distance covered from 2 min to 5 min = 0
Distance covered from 5 min to 8 min (at speed 2.20 m/s)
= (2.20 m/s) * (8 min - 5 min)
= (2.20 m/s) * (3 min)
= (2.20 m/s) * (3 * 60 s)
= 396 m
Time interval between 2 min and 8 min = 8 min - 2 min = 6 min = 360 s
Therefore, average speed between 2 min and 8 min = distance/time interval
= 396/360 m/s = 1.10 m/s
Velocity at 2 min = 0
Velocity at 8 min = 2.20 m/s
Change in velocity = 2.20 m/s - 0 = 2.20 m/s
Time interval = 8 min - 2 min = 6 min = 360 s
Avg acceleration = change in velocity/time period = 2.20/360 m/s^2
= 6.11 * 10^-3 m/s^2
——————————————————-
Can you calculate the rest?
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