Question

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Answer 1

Given,

the initial velocity = 0 m /s.

acceleration = 3.20 m / s^2

time = 32.8 s

According to laws of motion.

s = ut + 1/2 at ^2

s = 1/2 at²

s=1/2(3.20)(32.8)²

s= 1721.344 m

the distance traveled before takeoff is 1731.3m

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Answer 2

Answer:1.721km

Explanation:At first the plane is at rest so initial velocity of the plane is zero.It is given that acceleration a=3.2m/s2 and time t=32.8s.Applying Newton's laws of motion we have ,S=ut+1/2(at2).As u=0,the formula reduces to S=1/2(at2). Putting respective values we get,

S=1/2(3.2*(32.78*32.8))

S=1721.344m

S=1.721km