The Columbia Accident. On January 16, 2003, 81.7 seconds into the ascent, a piec

Question

The Columbia Accident. On January 16, 2003, 81.7 seconds into the ascent, a piece of foam, weighing about 1.7 pounds, and being roughly the size, shape and weight of a large loaf of bread, separated from the bipod ramp of the space shuttle Columbia's external fuel tank and impacted the leading edge of the shuttle's left wing. At that time the shuttle was traveling at approximately 1,568 mph (Mach 2.46) and was at an altitude of 66,000 feet. Based on film evidence, the foam traveled the 58-foot distance from the ramp to the wing in 0.16 seconds. Assuming a constant acceleration relative to the space shuttle of feet per second squared, the foam hit the wing at a speed of feet per second, or miles per hour. Unbeknownst to the astronauts and observers on earth, it struck the wing hard enough to cause the destruction of the Columbia two weeks later during re-entry on February 1, 2003.

Explanation / Answer

Under constant acceleration, s=1/2at2

So we get 58 = 1/2(a)(0.16)2

=> a = (58)(2)/(0.16)2 = 4531.25 ft/sec2

v = at = (4531.25)(0.16) = 725ft/sec

v=v0 + at

v0 = 0 so we get v = at

By definition, v = ds/dt = at

=> ds = atdt

Integrate this , we get s = at2/2 ( For constant acceleration)

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