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The position of a particle as a function of time
Chapter 3, Problem 21P(choose chapter or problem)
The position of a particle as a function of time is given by \(\vec{r}=(5.0 \hat{\imath}+4.0 \hat{\jmath}) t^{2} \mathrm{~m}\), where \(t\) is in seconds.
a. What is the particle's distance from the origin at \(t=0,2\), and \(5 \mathrm{~s}\)?
b. Find an expression for the particle's velocity \(\vec{v}\) as a function of time.
c. What is the particle's speed at \(t=0,2\), and \(5 \mathrm{~s}\)?
Equation Transcription:
Text Transcription:
^vec r=(5.0 ^hat i+4.0 ^hat j)t^2 m
t
t=0, 2, 5 s
^vec v
t=0, 2, 5 s
Questions & Answers
QUESTION:
The position of a particle as a function of time is given by \(\vec{r}=(5.0 \hat{\imath}+4.0 \hat{\jmath}) t^{2} \mathrm{~m}\), where \(t\) is in seconds.
a. What is the particle's distance from the origin at \(t=0,2\), and \(5 \mathrm{~s}\)?
b. Find an expression for the particle's velocity \(\vec{v}\) as a function of time.
c. What is the particle's speed at \(t=0,2\), and \(5 \mathrm{~s}\)?
Equation Transcription:
Text Transcription:
^vec r=(5.0 ^hat i+4.0 ^hat j)t^2 m
t
t=0, 2, 5 s
^vec v
t=0, 2, 5 s
ANSWER:
Step 1 of 3
(a)
We need to find out the particle distance.
The position vector,
\(\vec{r}=(5 \hat{i}+4 \hat{j}) t^{2}\)
The magnitude of \(\vec{r}\) is,
\(r=\sqrt{\left(5 t^{2}\right)^{2}+\left(4 t^{2}\right)^{2}}\)
\(r=6.40 t^{2} m\)
At time \(t=0 s\) particle distance is,
\(r=6.40 \times 0^{2}=0\)
At time \(t=2 \mathrm{~s}\) particle distance is,
\(r=6.40 \times 2^{2}=25.60 m\)
At time \(t=5 \mathrm{~s}\) particle distance is,
\(r=6.40 \times 5^{2}=160 m\)