The space-shuttle orbiter touches down at about . At its drag parachute deploys.

Question

The space-shuttle orbiter touches down at about . At its drag parachute deploys. At , the chute is jettisoned from the orbiter. If the deceleration in feet per second squared during the time that the chute is deployed is (speed in feet per second), determine the corresponding distance traveled by the orbiter. Assume no braking from its wheel brakes.

Explanation / Answer

when parachute opens v = 200ft/sec when it removed v = 35ft/sec ; a = dv/dt = -.0003v^2 dv /v^2 = -0.0003dt integrating both side with v from 200 -> v and t from 0 -> t we have v = 10000/(3t + 50 )ft/sec when v = 35 => t = 78.57 sec now dx = v dt => dx = [10000/(3t+50)] dt t from 0 -> 78.57sec integrating both side we have x = [10000/3] * ln((3t+50)/50) = 5809.85 ft is distance travelled during that time.

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