A plane has a minimum take off speed of 28.3 m/s. The length of the runway is 60

Question

A plane has a minimum take off speed of 28.3 m/s. The length of the runway is 600 m, and the angle at which the plane takes off is 20 degrees from the horizon. It continues to accelerate at 50% the take off acceleration (a2=0.5*a1) for 41 seconds. The pilot decides to land at an airport 2 km away, and the optimal landing speed is 21 m/s. The maximum runway deceleration is 0.56 m/s^2.

I found the minimum acceleration for take-off to be 0.67 m/s^2 using vf^2=vo^2+2as.

a. What is the cruising altitude during the flight? (height)

b. What is the cruising velocity?

c. If the runway is 400 m long, will the plane land safely?

d. What is the total horizontal displacement while the plane is in the air (not including runway and landing)?

Explanation / Answer

a)

during the flight

a = acceleration = 0.5 x 0.67 = 0.34 m/s2

t = time = 41 sec

vo = initial velocity = 0 m/s

y = altitude

using the equation

y = vo t + (0.5) a t2

y = (0) (41) + (0.5) (0.34) (41)2

y = 285.8 m

b)

cruising velocity is given as

vy = voy + at

vy = 0 + (0.34) (41) = 14 m/s

vx = 28.3 m/s

so v = sqrt(vy2 + vx2) = sqrt((28.3)2 + (14)2) = 31.6 m/s

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