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An airplane reaches its take off velocity of 60 m/s in 30 sec starting from the rest....
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The airplane starts from rest and reaches a velocity of 60 m/s in 30 s. The acceleration of the airplane or the rate of change of velocity is (60 - 0)/30 = 60/20 = 3 m/s^2.
The time taken to reach a velocity of 60 m/s starting from 40 m/s has to be determined. Assuming the average acceleration of the airplane is constant, use the relation v = u + a*t, where u is the initial velocity, v is the final velocity, a is the acceleration and t is the time taken.
Substituting the values given, 60 = 40 + 3*t
=> 20 = 3t
=> t = 20/3
=> t = 6.67 s
The airplane goes from 40 m/s to 60 m/s in 6.67 seconds.
Posted by justaguide on February 27, 2012 at 10:11 PM (Answer #1)
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