A rocket is 13 feet from a satellite when it begins accelerating away from the satellite at a constant rate of 6 feet per second squared. The distance, in feet, between the rocket and the satellite is modeled by [tex]P = 3t^2 + 13[/tex], where [tex]t[/tex] is the number of seconds since the rocket started accelerating.

Use the model to find and interpret the result when [tex]t = 4[/tex].

A. [tex]P(4) = 109[/tex] feet. The rocket is 109 feet away from the satellite after 4 seconds of acceleration.

B. [tex]P(4) = 61[/tex] feet. The rocket is 61 feet away from the satellite after 4 seconds of acceleration.

C. [tex]P(4) = 25[/tex] feet. The rocket is 25 feet away from the satellite after 4 seconds of acceleration.

D. [tex]P(4) = 97[/tex] feet. The rocket is 97 feet away from the satellite after 4 seconds of acceleration.

By substituting t = 4 into the given equation P=3t^2+13, we find that Option b) P(4) = 61 feet is correct. This means, the rocket is 61 feet away from the satellite after 4 seconds of acceleration.

Explanation:

To determine the distance between the rocket and the satellite after 4 seconds of acceleration, we substitute t = 4 into the given equation P=3t^2+13. So, P = 3*(4)^2 + 13 = 48 + 13 = 61 feet. This implies, Option b) P(4) = 61 feet is correct, denoting that the rocket is 61 feet away from the satellite after 4 seconds of acceleration. Remember that in such problems, t represents time elapsed and P is usually the position or distance from a reference point, here being the satellite.

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