Alight plane must reach a speed of 38 m/s for takeoff. how long a runway is needed if the (constant) acceleration is 2.8 m/s^2 ? express your answer using two significant figures.

The runway should be at least 0.27 km long.

Explanation:

Hi there!

The equations for position and velocity of the plane at a given time "t" are as follows:

x = x0 +v0 · t + 1/2 · a · t²

v = v0 + a · t

Where

x = position at time "t".

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity at time "t"

Let´s place the origin of the frame of reference at the point where the plane starts traveling to take off (x0 = 0). Since the plane starts from rest, v0 = 0.

First, using the equation of velocity, let´s find how much time it takes the plane to reach a velocity of 38 m/s.

v = v0 + a · t

38 m/s = 0 m/s + 2.8 m/s² · t

38 m/s / 2.8 m/s² = t

t = 14 s

Now, let´s find how much distance the plane travels in that time using the equation of position:

x = x0 +v0 · t + 1/2 · a · t²

x = 0 m + 0 m/s · t +1/2 · 2.8 m/s² · (14 s)²

x = 2.7 × 10² m

Then, the runway should be at least 0.27 km long.

Have a nice weekend!

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