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The change in velocity is the integral of acceleration with respect to time. Assuming a constant acceleration, then

v = integral [a dt] = a t + v0

The change in distance is the integral of velocity with respect to time:

s = integral (v dt) = integral [(at + v0) dt] = 1 /2 a t^2 + v0 t + s0

Since the airplane is taking off from standing still at the start of the runway, s0 = 0 and v0 = 0.

s = 1/2 a t^2

We know that the at the end of the runway (456 m), the velocity must be 195 mph or 87.2 m/s, and this is v = at for constant acceleration. Plugging this in, we get:

456 m = 1/2 (87.2 m/s) t

Solving for t we get t = 10.5 seconds

Plugging this back into the equation for s, then

456 m = 1/2 a (10.5 s)^2

Solving for a:

a = 8.27 m/s^2

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Q: An airplane must reach a speed of 195 miles hour to take off if the the runway is 456 meter long what is the minimum value of the acceleration that will allow the airplane to take off successfully?
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An airplane originally at rest on a runway accelerates uniformly at 6 meters per second for 12 seconds During this 12-second interval the airplane travels a distance of approximately?

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