Final answer:
Using the kinematic equation s = ut + (1/2)at^2, where u = 0 m/s (starting from rest), a = 5.20 m/s^2 (acceleration), and t = 41.7 s (time), we find that the airplane travels 4521.114 meters before takeoff.
Step-by-step explanation:
To determine the distance traveled before takeoff by an airplane that accelerates down a runway, we can use the kinematic equation that relates distance (s), acceleration (a), and time (t).
The kinematic equation for this scenario is:
s = ut + \(rac{1}{2}\)at^2
Where:
- u is the initial velocity (m/s), which is 0 m/s since the airplane starts from rest,
- a is the acceleration (m/s^2), which is given as 5.20 m/s^2,
- t is the time (s), which is given as 41.7 s,
- s is the distance (m) that we need to find.
Plugging the values into the equation, we get:
s = 0 \(\times\) 41.7 + \(rac{1}{2}\) \(\times\) 5.20 \(\times\) 41.7^2
s = 0 + \(rac{1}{2}\) \(\times\) 5.20 \(\times\) 1738.89
s = 2.6 \(\times\) 1738.89 m
s = 4521.114 m
Therefore, the airplane travels 4521.114 meters before it takes off the ground.