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Motion and rate of change derivatives. Attached is the question and where I am stuck, I would really appreciate the help :)

Motion and rate of change derivatives. Attached is the question and where I am stuck-example-1
Motion and rate of change derivatives. Attached is the question and where I am stuck-example-1
Motion and rate of change derivatives. Attached is the question and where I am stuck-example-2
User Fizor
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1 Answer

3 votes

Answer:

A. 3 < t < 9

Explanation:

You have successfully found the values of t where the motion turns around (t=3 and t=10). Your knowledge of odd-degree polynomials tells you that the value of s(t) is decreasing for t < 3 (because the leading coefficient is negative), and is also decreasing for t > 10.

That is, the value of s(t) is increasing between t=3 and t=10, which means the particle is moving to the right in that interval. Since the function is only defined on the interval [0, 9] the particle is moving to the right on the interval (3, 9).

_____

At t=3 and t=10, the particle is stationary, moving neither right nor left.

Motion and rate of change derivatives. Attached is the question and where I am stuck-example-1
User Tschale
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