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An airplane reaches its take off velocity of 60 m/s in 30 s starting from the rest....
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The airplane reaches its take off velocity of 60 m/s in 30 s starting from rest. The time spent in going from 40 m/s to 60 m/s has to be determined.
It is assumed that the velocity of the airplane increases at a constant rate or the acceleration is the same throughout the 30 s.
As the airplane starts from 0 and reaches 60 m/s in 30 s, the acceleration is (60 - 0)/30 = 2 m/s^2.
The time t required for its velocity to increase from 40 m/s to 60 m/s can be determined by solving 60 = 40 + 2*t
=> 20 = 2*t
=> t = 10
The airplane requires 10 s for its velocity to increase from 40 m/s to 60 m/s
Posted by justaguide on March 1, 2012 at 11:32 PM (Answer #2)
This answer will only work if the acceleration of the airplane is constant.
1. Find acceleration: The airplane starts at a velocity of 0m/s and reaches 60m/s in 30 seconds, meaning the acceleration is (60 m/s - 0m/s)/30 seconds. So the acceleartion is 2 m/s^2.
2. Apply this acceleration into the second equation:
Let the time equals t so,
solving for t gives us: 2t=20, so t=10.
Thus, the answer is 10 seconds.
Posted by dweedledum360 on March 9, 2012 at 11:14 AM (Answer #3)
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