A plane accelerates from rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long. Assume that the plane reaches the required takeoff velocity at the end of the runway. What is the distance dlast traveled by the plane in the last second before taking off?

132 meters

Explanation:

Given:

x₀ = 0 m

x = 1800 m

v₀ = 0 m/s

a = 5.00 m/s²

First, find the time to reach the end of the runway:

x = x₀ + v₀ t + ½ at²

1800 = 0 + 0 + ½ (5) t²

t = √720

Now find the position one second before:

x = x₀ + v₀ t + ½ at²

x = 0 + 0 + ½ (5) (√720 - 1) ²

x ≈ 1668

The distance from the end of the runway is:

1800 - 1668 = 132

So the plane travels 132 meters in the last second.