Answer:
The speed of the plane at the end of the runway is 105 m/s.
Step-by-step explanation:
Hi there!
The equations of position and velocity in function of time of an object moving in a straight line with constant acceleration are the following:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where:
x = position of the plane at time t.
x0 = initial position.
v0 = initial velocity.
t = time.
a = acceleration.
v = velocity of the plane at time t.
If we place the origin of the frame of reference at the starting point, then x0 = 0. Since the plane starts from rest, v0 = 0. Then, the equation gets reduced to this:
x = 1/2 · a · t²
Let's find how much time it takes the plane to travel 500 m:
500 m = 1/2 · 11.09 m/s² · t²
2 · 500 m / 11.09 m/s² = t²
t = 9.50 s
Now, let's use the equation of velocity to find the speed of the plane at t = 9.50 s (the time at which the plane is at the end of the runway):
v = v0 + a · t (v0 = 0)
v = a · t
v = 11.09 m/s² · 9.50 s
v = 105 m/s
The speed of the plane at the end of the runway is 105 m/s.