An airplane starts from rest and accelerates at 11.09 m/s2. What is its speed at the end of a 500-m runway?

The speed of the plane at the end of the runway is 105 m/s.

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.