The position of a particle is given by the function x =

Chapter 2, Problem 31P

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QUESTION:

Problem 31P

The position of a particle is given by the function x = (2t 3 − 9t 2 + 12) m, where t is in s.

a. At what time or times is vx = 0 m/s?

b. What are the particle’s position and its acceleration at this lime(s)?

Questions & Answers

QUESTION:

Problem 31P

The position of a particle is given by the function x = (2t 3 − 9t 2 + 12) m, where t is in s.

a. At what time or times is vx = 0 m/s?

b. What are the particle’s position and its acceleration at this lime(s)?

ANSWER:

Step 1 of 3

We have at what time or times is \(v_{x}=0 \mathrm{~m} / \mathrm{s}\) for a particle whose position is given by the function \(x=2 t^{3}-9 t^{2}+12\)

The velocity function of the particle is given by the expression.

                                               \(v_{x} =\frac{d x}{d t}\)

                                               \(=6 t^{2}-18 t\)

Now, the time at which \(v_{x}=0 \mathrm{~m} / \mathrm{s}\) can be found by solving the equation for \(v_{x}\) with \(v_{x}=0\)

Thus,

                                                 \(6 t^{2}-18 t=0\)

                                                 \(6 t(t-3)=0\)

Hence,

At \(t=0 \mathrm{~s}\) and \(\mathrm{t}=3 \mathrm{~s}, v_{x}=0\)

Therefore, at times \(t=0 \mathrm{~s}\) and \(\mathrm{t}=3 \mathrm{~s}, v_{x}=0\).

 

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