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.

A plane accelerates from rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long. Assume that the plane reaches the required takeoff velocity at the end of the runway. What is the time tTO needed to take off?

What is the distance dfirst traveled by the plane in the first second of its run?

What is the distance dlast traveled by the plane in the last second before taking off?

Explanation / Answer

a)
use:
d = vi*t+0.5*a*t^2
1800 = 0*t + 0.5*5*t^2
1800 = 0.5*5*t^2
t = 26.8 seconds

b)
for 1st seconds
d = vi*t+0.5*a*t^2
d = 0*1 + 0.5*5*1^2
d = 0.5*5*1
d = 2.5 m

c)
To find distance in last seconds. Lets find distance travelled in 25.8 seconds and subtract it with total length to get length in last second
at t = 25.8 s
d = vi*t+0.5*a*t^2
d = 0*1 + 0.5*5*(25.8)^2
d = 1664 m
In last second, it covers : 1800 - 1664 = 136 m

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