An airplane starts from rest, 6050 ft down a runway at uniform accelerationthen takes off with a speed of 150mi / h. It then climbs in a straight line with a uniform acceleration of 2 ft/s^ 2 until it reaches a constant speed of 195mi / h. How far has the plane traveled when it reaches this constant speed?

Given Information:

distance = s₁ = 6050 ft

velocity = v₁ = 0 mi/hr

velocity = v₂ = 150 mi/hr

velocity = v₃ = 195 mi/hr

Acceleration = a = 2 ft/s²

Required Information:

distance = s₂ = ?

Answer:

distance = s₂ = 14,399 ft

Explanation:

We know from the equations of motion,

v₃² = v₂² + 2a (s₂ - s₁)

We want to find out the distance s₂

2a (s₂ - s₁) = v₃² - v₂²

s₂ - s₁ = (v₃² - v₂²) / 2a

s₂ = (v₃² - v₂²) / 2a + s₁

First convert given velocities from mi/hr to ft/s

1 mile has 5280 feet and 1 hour has 3600 seconds

velocity = v₂ = 150 * (5280/3600) = 220 ft/s

velocity = v₃ = 195 * (5280/3600) = 286 ft/s

s₂ = (v₃² - v₂²) / 2a + s₁

s₂ = (286² - 220²) / 2*2 + 6050

s₂ = 33396/4 + 6050

s₂ = 8349 + 6050

s₂ = 14,399 ft

Therefore, the plane would have traveled a distance of 14,399 ft when it reaches a constant speed of 286 ft/s