An airplane starts from rest and accelerates at 5.99 m/s2. What is its speed at the end of a 500-m runway?

Question

anonymous · Answer

V(final)^2 = V(initial)^2 +2*a*x  

you can use this equation.

anonymous · Answer

^

vf^2=vi^2+2ad

vf^2=2ad

anonymous · Answer

Supposing it travels in a straight line, one could use the kinematic equation for uniformly accelerated motion.
$$FinalVelocity^{2} = InitialVelocity^{2} + 2 * Acceleration * Distance$$
Which, provided that you start from rest, and travel 500m, becomes:
$$FinalVelocity^{2} = 2 * 5.99 \frac{ m }{ s^{2} } * 500 m$$
$$FinalVelocity^{2} = 5990 \frac{ m^{2} }{ s^{2} }$$
$$FinalVelocity = \begin{matrix}+ \ -\end{matrix} 77.40 \frac{ m }{ s }$$
We assume that the plane does not fly backwards, so:
$$FinalVelocity = + 77.40 \frac{ m }{ s }$$

goformit100 · Answer

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