Vo = 0
V = 34.72 m/s.
a = 5.6 m/s^2
A. V = Vo + a*t = 34.72
0 + 5.6t = 34.72
t = 6.2 s.
Length = Vo*t + 0.5a*t^2
Length = 0 + 2.8*6.2^2 = 107.6 m.
B. V^2 = Vo^2 + 2a*L = 34.72^2 = 1205.5
0 + 2a*75 = 1205.5
150a = 1205.5
a = 8.04 m/s^2.
C. V = Vo + a*t = 34.72 m/s.
0 + 8.04t = 34.72
t = 4.32 s.
D. Vo=500km/h = 500,000m/3600s=138.9 m/s
= Initial velocity.
V = Vo + a*t = 0
138.9 - 4.5t = 0
4.5t = 138.9
t = 30.86 s.
E. L=1000yds * 3ft/yd * 1m/3.3Ft=909.1 m
d=(V^2-Vo^2)/2a = (0-138.9^2)/-9=2144 m.
= Stopping distance.
No, the plane will not be able to land safely.
A jet plane has a takeoff speed of Vto=125km/h (34.72 m/s) and it's engines can power it to accelerate with an average acceleration of 5.6 m/s^2.
A.) what length of runway will it need to take off safely?
B.) suppose it had an available runway length of only 75m. What minimum constant acceleration would it need to take off safely?
C.) supposing the runway length and acceleration in question b, how long will the plane take to take off?
D.) upon arriving at its destination the plane lands with a speed of 500 km/h and it's brakes can cause it to accelerate at -4.5 m/s^2 without severely compromising the comfort of the passengers. How long will the airplane take to come to rest?
E.) suppose the runway is 1000 yards long. Will the plane be able to land safely?
1 answer