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An initially stationary object experiences an acceleration of 6 m/s^2 for a time of 15s. How far will it travel during that time?x = xo + vo t + ½ a t^2 g = 9.8 m/s2 (downward)

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ANSWER

[tex]675m[/tex]

EXPLANATION

Parameters given:

Initial velocity, v0 = 0 m/s

Acceleration, a = 6 m/s^2

Time, t = 15s

To find the distance traveled during that time, we have to apply one of Newton's equations of motion:

[tex]x=ut+\frac{1}{2}at^2[/tex]

Therefore, we have:

[tex]\begin{gathered} x=(0\cdot15)+(\frac{1}{2}\cdot6\cdot15^2) \\ x=0+675 \\ x=675m \end{gathered}[/tex]

That is the distance traveled during that time.

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