A plane starting at rest at the south end of a runway undergoes a constant acceleration of 1.6 m/s/s for a distance of 1600m before takeoff. What is the time required for takeoff? What is the plane's velocity at takeoff?

Question

Anne Proctor

Physics

21 July, 17:41

A plane starting at rest at the south end of a runway undergoes a constant acceleration of 1.6 m/s/s for a distance of 1600m before takeoff. What is the time required for takeoff? What is the plane's velocity at takeoff?

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Answers (1)

Know the Answer?

Barney · Answer 1

V^2 = Vi^2 + 2ax

Plugging in the numbers for the variables gives me

v^2 = 0^2 + 2 x 1.6 x 1600

so

v^2 = 5120

sq (5120

V = 71.554 m/s

For part B, all I did was:

t = Vf - Vi / A

Plugging in the numbers for the variables gives me

t = 71.554 - 0 / 1.6

so

t = 44.72 seconds for the plane to take off