An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts offthe ground. Determine the distance traveled before takeoff
Question
An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts offthe ground. Determine the distance traveled before takeoff
Solution
To solve this problem, we can use the formula for distance in uniformly accelerated motion:
d = vit + 0.5at^2
where:
- d is the distance,
- vi is the initial velocity,
- a is the acceleration, and
- t is the time.
In this case, the airplane starts from rest, so the initial velocity (vi) is 0. The acceleration (a) is 3.20 m/s^2 and the time (t) is 32.8 s. Substituting these values into the formula gives:
d = 0*(32.8 s) + 0.5*(3.20 m/s^2)(32.8 s)^2 d = 0 + 0.53.20 m/s^2*(1075.84 s^2) d = 0 + 1721.344 m d = 1721.344 m
So, the airplane traveled approximately 1721.34 meters before takeoff.
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