How Fast is an Airplane Moving at Takeoff with Constant Acceleration?

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SUMMARY

An airplane travels 277 meters down the runway before takeoff, achieving this in 8.00 seconds under constant acceleration. The average velocity of 34.6 m/s is incorrect for determining takeoff speed. To find the velocity at takeoff, one must first calculate the constant acceleration using the equation x(t) = 1/2 a t^2, and then apply v(t) = a*t. The average velocity is half of the final velocity when acceleration is constant.

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airplane down the runway...

An airplane travels 277m down the runway before taking off.
Assuming that it has constant acceleration, if it starts from rest and becomes airborne in 8.00 seconds, how fast m/s is it moving at takeoff?

am i missing something here? 277/8seconds =34.6m/s?
 
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yes, I'm afraid you are. What you have calculated is it's average velocity, the total displacement divided by the total time. To find the velocity at take off, you first have to find what that constant acceleration is, then see how fast it's going after 8 seconds of accelerating at that rate.

you need these two equations:

x(t) = 1/2 a t^2 (assuming it starts from rest at the origin)

and

v(t) = a*t (assuming it starts from rest)
 
It's easier than that, with constant acceleration, average velocity is 1/2 the final velocity.
 

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