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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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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?

The formula for distance covered during uniform acceleration isd = 1/2 * (vf + vi) * t (1)Time, t, is given; initial velocity, vi, is 0; but final velocity, vf, is unknown and must be computed from given information. Knowing the rate of acceleration, initial velocity and time, The final velocity may be computed using the formula for average acceleration (actual acceleration under uniform motion) which isa = (vf - vi) / t (2)Rewriting to solve for vf with vi = 0 we havevf = a * tvf = 6m/s2 * 12svf = 72m/sPlugging this value into equation (1) with the other given values we haved = 1/2 * (72m/s + 0 m/s) * 12sd = 432mSo the airplane will travel 432m from rest in 12 seconds under 6m/s uniform acceleration.


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