Arturo O. answered • 09/09/17

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The plane needs to get from 0 m/s to 16 m/s at a constant acceleration of 3.2 m/s

^{2}. Solve this using the kinematic equationv

_{f}^{2}- v_{i}^{2}= 2adv

_{f}= final speed = 16 m/sv

_{i}= initial speed = 0 m/sa = constant acceleration = 3.2 m/s

^{2}d = distance over which the acceleration occurs

Solve for d. If d ≤ 50 m, the plane can take off.

d = (v

_{f}^{2}- v_{i}^{2})/(2a) = (16^{2}- 0^{2})/[2(3.2)] m = 40 md = 40 m < 50 m, so there is enough runway for takeoff.