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The 226,000-lb space shuttle orbiter touches down at about 221 mi/hr. At 181 mi/hr, its drag parachute deploys. At 28 mi/hr, the chute is jettisoned from the orbiter. If the deceleration in feet per second squared during the time that the chute is deployed is \(-0.00025v^2\) (speed \(v\) in feet per second), determine the corresponding distance \(s\) traveled by the orbiter. Assume no braking from its wheel brakes.

Answer :

The corresponding distance traveled by the orbiter is 224,761.8 miles

To calculate the distance traveled by the orbiter while the drag parachute is deployed, we need to first determine the time it takes for the speed to decrease from 221 mi/hr to 28 mi/hr.

We can use the formula:

t = (v2 - v1) / a

where t is the time, v2 is the initial speed, v1 is the final speed, and a is the deceleration.

Plugging in the given values, we find that t = 1,648 seconds.

The corresponding distance traveled by the orbiter can be calculated using the formula:

s = v2 * t + 0.5 * a * t2

Substituting the values, we find that s = 224,761.8 miles.

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