Section 2.23  Interactive checkpoint: passenger jet  
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A passenger jet lands on a runway with a velocity of 71.5 m/s. Once it touches down, it accelerates at a constant rate of −3.17 m/s2.
How far does the plane travel down the runway before its velocity is decreased to 2.00 m/s, its taxi speed to the landing gate?
 
 

Answer:

Δx = m

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Diagram  [Hide]

Variables  [Hide]

Fill in the known values.

Variable

Value
acceleration

a = m/s2

initial velocity

vi = m/s

final velocity

vf = m/s

distanced traveled

Δx

time elapsed

t

I think I'm done. I need help. Show the correct values.
 
Strategy  [Hide]
  1. Select an equation that contains only the variables you are given, and the one you want to calculate. Avoid equations with two unknowns.
Physics principles and equations  [Hide]

vf = vi + at

Δx = vit + ½at2

vf2 = vi2 + 2aΔx

Δx = ½(vi + vf)t

Step-by-step solution   [Hide]
3

Select the equation that will help you calculate the distance traveled.

Solve for distance traveled

vf2 – vi2 = 2aΔx

Δx = (vf2 – vi2)/ 2a

Enter values

Δx = [( m/s)2 – ( m/s)2]/ 2( m/s2)

Evaluate

Square the velocities and subtract.

Δx = ( m2/s2)/ 2(–3.17 m/s2)

Divide

Δx = m
 
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