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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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gretjxfj | Student, Undergraduate | (Level 1) Honors

Posted March 1, 2012 at 7:23 PM via web

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An airplane reaches its take off velocity of 60 m/s in 30 s starting from the rest. Find the time it spends in going from 40 m/s to 60 m/s .

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted March 1, 2012 at 11:32 PM (Answer #2)

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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

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dweedledum360 | Student, Grade 9 | (Level 1) eNoter

Posted March 9, 2012 at 11:14 AM (Answer #3)

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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,

2m/s^2=(60m/s-40m/s)/t

solving for t gives us: 2t=20, so t=10.

Thus, the answer is 10 seconds.

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