Final answer:
The distance traveled by an airplane before takeoff, given that it starts from rest and accelerates at 3.50 m/s² for 34.5 s, is calculated using the kinematic equation for distance. After substituting the known values, the result is 2082.94 meters.
Step-by-step explanation:
The student is asking how to use the kinematic equation for distance to find out how far an airplane travels before takeoff if it starts from rest and accelerates at a constant rate over a period of time. This is a physics question involving kinematic equations, specifically the one for calculating distance traveled when the initial velocity, acceleration, and time are known.
To solve the problem, we use the equation d = v0t + ½at², where:
- v0 is the initial velocity
- d is the displacement or distance traveled
- t is the time
- a is the acceleration
In this scenario, the airplane starts from rest, so v0 = 0, it accelerates at a = 3.50 m/s², and this occurs over a time t = 34.5 s. Substituting these values into the equation gives:
d = 0 × 34.5 s + ½(3.50 m/s²)(34.5 s)²
Calculating this yields:
d = ½(3.50 m/s²)(1190.25 s²)
Which simplifies to:
d = ½(4165.875 m)
And further simplifies to:
d = 2082.9375 m
Therefore, the distance traveled by the airplane before takeoff is 2082.94 meters.