Question

helpppppppp
On the Moon, a falling object falls just 2.65 feet in the first second after being dropped. Each second it falls 5.3 feet farther than in the previous second. How far would an object fall in the first ten seconds after being dropped?

Answer 1

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Answer 2

I'm fairly sure I confused you with the other posts so here's a full solution from start to finish.

The acceleration is actually going to be 5.3 ft/s^2 since that's the rate of change in velocity with respect to time. To calculate the distance, you're going to need to use the kinematic equation:
\[d=V_it+\frac{1}{2}at^2\]Vi = 0 ft/s (this is the initial velocity)
t = 10 seconds
a = 5.3 ft/s^2

Plugging in the values you get:
\[d=(0ft/s)(10s)+\frac{1}{2}(5.3ft/s^2)(10s)^2=265ft\]That's all there is to it.