Question

What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long. assume that the plane reaches the required takeoff velocity at the end of the runway. what is the time tto needed to take off?

Answer 1

The plane gains 2400% of its takeoff velocity when it reaches the midpoint of the runway.

Step-by-step explanation:

To find the percentage of the takeoff velocity gained at the midpoint of the runway, we need to calculate the time it takes for the plane to reach the midpoint.

First, we can use the equation v = v0 + at to find the time it takes for the plane to take off. Since the initial velocity (v0) is 0 m/s and the final velocity (v) is 60 m/s, and the acceleration (a) is constant at 5.00 m/s², we can rearrange the equation to solve for t.

Using the formula and solving for t, we get: t = (v - v0) / a. Substituting the values, we have: t = (60 - 0) / 5.00 = 12 s. So, it takes 12 seconds for the plane to take off.

Since the midpoint is at half the distance of the runway, which is 1800 m / 2 = 900 m, we can use the formula d = v0t + (1/2)at² to find the velocity at the midpoint.

Substituting the values, we have: d = (0)(12) + (1/2)(5.00)(12²) = 1,440 m.

Therefore, the plane gains a velocity of 1,440 m/s when it reaches the midpoint of the runway. To find the percentage, we divide the gained velocity by the takeoff velocity and multiply by 100.

Using the formula, we have: Percentage gained = (Gained velocity / Takeoff velocity) × 100 = (1,440 / 60) × 100 = 2400%. So, the plane gains 2400% of its takeoff velocity when it reaches the midpoint of the runway.

Answer 2

When is at the end of the runway the velocity of the plane is given by the equation
`vf^(2)=0+2*a*s` where s=1800 m is the runway length. Thus

`vf^(2)=2*5*1800=18000 (m/s)^(2)`

`vf =134.164 (m/s)`

At half runway the velocity of the plane is

`v^(2)=2*5* (1800)/(2)=9000 ( (m)/(s) )^(2)`

`v= √(9000)=94.87 ( (m)/(s))`

Therefore at midpoint of runway the percentage of takeoff velocity is
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`P= (v)/(vf)= (94.87)/(134.164)=0.707`