Asked by talal
A fight aircraft needs to emergency land due to engine
breakdown. The pilot can control the approach velocity to 250
km/hr when the aircraft tires touch the ground. The best
deceleration of its braking system is 10 m/s2
. How long will the
runway is required for a safe landing?
breakdown. The pilot can control the approach velocity to 250
km/hr when the aircraft tires touch the ground. The best
deceleration of its braking system is 10 m/s2
. How long will the
runway is required for a safe landing?
Answers
Answered by
Damon
250 (1000/3600) = 69.4 m/s
v = Vi + a t
0 = 69.4 -10t
t = 6.94 seconds to stop
d = (1/2) a t^2
= (1/2)(10)(6.94^2)
= 241 meters
v = Vi + a t
0 = 69.4 -10t
t = 6.94 seconds to stop
d = (1/2) a t^2
= (1/2)(10)(6.94^2)
= 241 meters
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