Question

A plane is traveling at a velocity of 90 m/s. It accelerates at a constant rate of 1.5 m/s2 until its velocity reaches 500 m/s. What distance did the plane cover while it was accelerating?

Answer 1

Recall that

v ² - v₀² = 2 a ∆x

where

v₀ = initial velocity

v = final velocity

a = acceleration

∆x = distance

Then

(500 m/s)² - (90 m/s)² = 2 (1.5 m/s²) ∆x

∆x = ((500 m/s)² - (90 m/s)²) / (2 (1.5 m/s²))

∆x ≈ 80,633 m