To take off from the ground, an airplane must reach a sufficiently high speed. T

Question

To take off from the ground, an airplane must reach a sufficiently high speed. The velocity required for the takeoff, the takeoff velocity, depends on several factors, including the weight of the aircraft and the wind velocity.
1)A plane accelerates from rest at a constant rate of 5.00 m/s sqaure along a runway that is 1800m long. Assume that the plane reaches the required takeoff velocity at the end of the runway. What is the time t_{TO} needed to take off?
2)What is the speed v_{TO} of the plane as it takes off?

Explanation / Answer

Given that the initial velocity is u = 0 m/s Acceleration is a = 5.00 m/s^2 Total distance is S = 1800 m -------------------------------------------------------- (1) From the equation of motion displacement is S = ut + (1/2)at^2 1800 m = 0 + (1/2)(5.00 m/s^2)t^2 t ^2 = 2*1800m /(5.00 m/s^2) = 720 s^2 t = 26.832 s (2) From the equation of kinematics final velocity v = u + at = 0 + (5.00 m/s^2)(26.832s) = 134.16 m/s

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