Question

The 243000-lb space-shuttle orbiter touches down at about 236 mi/hr. The drag chute is deployed at 189 mi/hr, the wheel brakes are applied at 101 mi/hr until wheelstop, and the drag chute is jettisoned at 35 mi/hr. If the drag chute results in a deceleration of -0.000200v2 (in feet per second squared when the speed v is in feet per second) and the wheel brakes cause a constant deceleration of 3.5 ft/sec2, determine the distance s traveled from 189 mi/hr to wheelstop.

Answer 1

Answer:

5156.37 ft

Explanation:

Given data:

weight ( W ) = 243,000 Ib

Motion of shuttle ; from 189 mi/hr to 101 mi/hr

dv/dt = -0.0002 V^2

I/v * dv/dt = -0.0002 ds

Convert mi/hr to ft/s ( 1 mi/hr = 1.467 ft/s)

189 mi/hr = 277.263 ft/s

101 mi/hr = 148.167 ft/s

After Integrating

In ( 148.167 / 277.263 ) = -0.0002 ( S1 - S2 )

S1 - S2 = -0.627 / -0.0002

S1 - S2 = 3135 ft/s

Now from 101 mi/hr to 35 mi/hr

dv/dt = ( - 0.0002 V^2 + 3.5 )

ds =  V*dv / ( -0.0002 v^2 - 3.5 )

given :  35 mi/hr = 51.345 ft/s

             101 mi/hr = 148.167 ft/s

Integrate

S3 - S2 = - In( 0.0002 v^2 + 3.5 ) / 0.0002 * 2 ]

            = 1644.75 ft/s

S4 - S3 = 376.62 ft/s

attached below is the remaining part of the solution

Total distance travelled = 3135 + 1644.75 + 376.62 = 5156.37 ft