Solution Found!
The position of a particle is given by the function x =
Chapter 2, Problem 31P(choose chapter or problem)
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 \(\mathrm{t}=3 \mathrm{~s}, v_{x}=0\).