A Boeing 747 airplane is flying at the elevation of 2,000 km with the constant s

Question

A Boeing 747 airplane is flying at the elevation of 2,000 km with the constant speed of 250 m/s in the horizontal direction at 200 km away from the Baton Rouge airport (BTR). The airplane just started reducing its horizontal speed to the landing speed of 150 m/s before it starts the landing gear. How long does the airplane take to get the landing speed, if it slows down with the constant acceleration of -0.2 m/s^2 in the horizontal direction? While the airplane is slowing down in the problem a) above, how long horizontal distance docs this airplane fly? This calculation will give the distance from the airport where the airplane starts preparing landing. The airplane lowers its elevation by 5 km during the airplane is slowing down in the problem b) above. What is the acceleration in the vertical direction for this time period? Write the acceleration in this period in a vector form. In addition, what is the angle of this acceleration. What is the velocity of the airplane at this moment, just before it starts landing gear? The length of the runway of the Baton Rouge airport (BTR) is 2 km or 2000 m. If the airplane slows down with a constant acceleration, what would be the minimum acceleration in horizontal direction needed for this airplane to stop within the runway?

Explanation / Answer

Given,

elevation = 2000 km

initial speed = 250 m/s

distance = 200 km

final speed = 150 m/s

v = u + at

150 = 250 - 0.2 * t

t = 500 sec

time airplane will take to get the landing speed = 500 sec

v^2 = u^2 + 2as

150^2 = 250^2 - 2 * 0.2 * s

s = 100000 m

distance airplane will fly = 100000 m

s = 0.5 * at^2

5000 = 0.5 * a * 500^2

vertical acceleration = 0.04 m/s^2

acceleration in vector form = (-0.2i + 0.04j) m/s^2

angle = tan^-1(0.04 / 0.2)

angle = 11.31 degree

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