A 6.4 ton military fighter must land on a flight strip. If the plane lands at a speed of 368 mph, and the coefficient of kinetic friction is 0.72 between the wheels and the ground. If the runway must be at least 150% as long as the shortest possible landing distance of the fighter, what is the shortest distance that the runway can be constructed?

Question

HELP PLEASE I HAVE NO IDEA!

Responses

Math - bobpursley, Tuesday, June 2, 2009 at 9:41pm
Convert mph to m/s. It is about 165m/s

mg*mu*distance=1/2 vi^2 * m

Notice mass m divides out.
distance=1/2*1/mu*vi^2

Math - Chelsea, Tuesday, June 2, 2009 at 9:49pm
what is mu and vi? the last part 1/2*1/mu*vi^2

Answer 1

Assume that the deceleration depends entirely on friction between wheels and runway, i.e. ignore turbine in reverse, and the slippage of the brake.

Then the standard kinematics equations apply:
v²-u² = 2aS

v=final velocity = 0 (at rest)
u=initial velocity = 368 mph=368*44/30 fps
a=kinetic frictional force devided by mass
=umg/m (m cancels out here)
=ug
=-0.72*32.3 f/s/s (negative for deceleration)
=-23.184 f/s/s

Substituting in values,

S=539.73²/(2*23.184)
=6286.6 ft

Apply a safety factor of 1.5 to the calculated length, so length of runway to be constructed
= 6286.6*1.5 ft.
= 9424 ft.

Answer 2

Thank you so much for explaining :)