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. a. What is the time tTO needed to take off?b. What is the distance traveled by the plane in the last second before taking off?

26.83secs

Explanation: s= UT + 1/2at^2

U=0

S= 1/2at^2

1800=1/2*5t^2

3600=5t^2

3600/5=t^2

720=t^2

t=√720

t=26.83secs

b would be the answer

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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