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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?

Respuesta :

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

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