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

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An airplane travels 277 meters down the runway and takes off in 8 seconds, prompting a calculation of its takeoff speed. The initial calculation of 34.6 m/s represents average velocity, not the final velocity at takeoff. To determine the takeoff speed, one must first calculate the constant acceleration using the equation x(t) = 1/2 a t^2. The final velocity can then be found using v(t) = a*t, with the understanding that average velocity equals half the final velocity under constant acceleration. Accurate calculations are essential for determining the actual speed at takeoff.
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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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