An airplane from rest accelerates on a runway at 5.50 m/s2 for 20.25 s until it finallytakes off the ground. What is the distance covered before takeoff?
Question
An airplane from rest accelerates on a runway at 5.50 m/s² for 20.25 s until it finally takes off the ground. What is the distance covered before takeoff?
Solution
To solve this problem, we can use the equation of motion:
d = ut + 0.5at^2
where: d = distance covered u = initial velocity a = acceleration t = time
Given in the problem: u = 0 (since the airplane starts from rest) a = 5.50 m/s^2 t = 20.25 s
Substituting these values into the equation, we get:
d = 0*(20.25) + 0.5*(5.50)*(20.25)^2
Solving this, we find:
d = 0 + 0.55.50(20.25)^2 d = 0 + 0.55.50410.0625 d = 0 + 1125.171875 d = 1125.17 m
So, the airplane covers a distance of approximately 1125.17 meters before takeoff.
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