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The position of a squirrel running in a park is given by r⃗ =[(0.280m/s)t+(0.0360m/s2)t2]i^+ (0.0190m/s3)t3j^.

At 5.01 s , how far is the squirrel from its initial position?

At 5.01 s , what is the magnitude of the squirrel's velocity?

At 5.01 s , what is the direction (in degrees counterclockwise from +x-axis) of the squirrel's velocity?

Express your answer to three significant figures and include the appropriate units.

Guest Sep 8, 2015
edited by Guest  Sep 8, 2015

Best Answer 

 #1
avatar+18712 
+20

The position of a squirrel running in a park is given by r⃗   = [(0.280m/s)t+(0.0360m/s2)t2]i+ [ (0.0190m/s3)t3 ] j

a) At 5.01 s , how far is the squirrel from its initial position?

b) At 5.01 s , what is the magnitude of the squirrel's velocity?

c) At 5.01 s , what is the direction (in degrees counterclockwise from +x-axis) of the squirrel's velocity?

Express your answer to three significant figures and include the appropriate units.

 

\(\begin{array}{rcl} \vec{r} &=& \left[ 0.280\ \frac{m}{s}\cdot t+0.0360\ \frac{m}{s^2}\cdot t^2 \right] \vec{i}+ \left[ 0.0190\ \frac{m}{s^3}\cdot t^3 \right] \vec{j}\\ \\ \vec{r} &=& \dbinom{x(t)}{y(t)}= \begin{pmatrix} 0.280\ \frac{m}{s}\cdot t+0.0360\ \frac{m}{s^2}\cdot t^2 \\ 0.0190\ \frac{m}{s^3}\cdot t^3 \end{pmatrix}\\\\\\ \text{a) } t = 5.01~s\\ x(5.01~s)&=&0.280\ \frac{m}{s}\cdot (5.01~s)+0.0360\ \frac{m}{s^2}\cdot (5.01~s)^2 = 2.30640360000~m\\ y(5.01~s)&=&0.0190\ \frac{m}{s^3}\cdot (5.01~s)^3= 2.38927851900~m\\ r(5.01~s)&=&\sqrt{2.30640360000^2~m^2+2.38927851900^2~m^2} = 3.32086576173~m \end{array}\)

 

The squirrel is 3.32 m from its initial position(0,0)

 

\(\begin{array}{rcl} \vec{v} &=& \dbinom{v_x}{v_y} = \begin{pmatrix} \frac{d\ x(t)}{dt} \\ \frac{d\ y(t)}{dt} \end{pmatrix} = \begin{pmatrix} 0.280\ \frac{m}{s}+2\cdot 0.0360\ \frac{m}{s^2}\cdot t \\3\cdot 0.0190\ \frac{m}{s^3}\cdot t^2 \end{pmatrix}\\\\ \text{b) } t = 5.01~s\\ v_x(5.01~s)&=&0.280\ \frac{m}{s}+2\cdot 0.0360\ \frac{m}{s^2}\cdot (5.01~s) = 0.64072~\frac{m}{s}\\ v_y(5.01~s)&=&3\cdot 0.0190\ \frac{m}{s^3}\cdot (5.01~s)^2= 1.4307057~\frac{m}{s}\\ v(5.01~s)&=&\sqrt{0.64072^2~(\frac{m}{s})^2+1.4307057^2~({\frac{m}{s}})^2} = 1.56762269645~\frac{m}{s}\\ \end{array}\)

 

The magnitude of the squirrel's velocity is 1.57 m/s

 

\(\begin{array}{rcl} \text{c) direction } &=& \arctan{ \left( \frac{v_y}{v_x} \right) } \\ &=& \arctan{ \left( \frac{1.4307057~\frac{m}{s}}{0.64072~\frac{m}{s}} \right) } \\ &=& \arctan{ \left(2.23296556998 \right) } \\ &=& 65.8754973481^{\circ} \end{array}\)

 

The direction (in degrees counterclockwise from +x-axis) of the squirrel's velocity is 65.9 degrees

 

laugh

heureka  Sep 8, 2015
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1+0 Answers

 #1
avatar+18712 
+20
Best Answer

The position of a squirrel running in a park is given by r⃗   = [(0.280m/s)t+(0.0360m/s2)t2]i+ [ (0.0190m/s3)t3 ] j

a) At 5.01 s , how far is the squirrel from its initial position?

b) At 5.01 s , what is the magnitude of the squirrel's velocity?

c) At 5.01 s , what is the direction (in degrees counterclockwise from +x-axis) of the squirrel's velocity?

Express your answer to three significant figures and include the appropriate units.

 

\(\begin{array}{rcl} \vec{r} &=& \left[ 0.280\ \frac{m}{s}\cdot t+0.0360\ \frac{m}{s^2}\cdot t^2 \right] \vec{i}+ \left[ 0.0190\ \frac{m}{s^3}\cdot t^3 \right] \vec{j}\\ \\ \vec{r} &=& \dbinom{x(t)}{y(t)}= \begin{pmatrix} 0.280\ \frac{m}{s}\cdot t+0.0360\ \frac{m}{s^2}\cdot t^2 \\ 0.0190\ \frac{m}{s^3}\cdot t^3 \end{pmatrix}\\\\\\ \text{a) } t = 5.01~s\\ x(5.01~s)&=&0.280\ \frac{m}{s}\cdot (5.01~s)+0.0360\ \frac{m}{s^2}\cdot (5.01~s)^2 = 2.30640360000~m\\ y(5.01~s)&=&0.0190\ \frac{m}{s^3}\cdot (5.01~s)^3= 2.38927851900~m\\ r(5.01~s)&=&\sqrt{2.30640360000^2~m^2+2.38927851900^2~m^2} = 3.32086576173~m \end{array}\)

 

The squirrel is 3.32 m from its initial position(0,0)

 

\(\begin{array}{rcl} \vec{v} &=& \dbinom{v_x}{v_y} = \begin{pmatrix} \frac{d\ x(t)}{dt} \\ \frac{d\ y(t)}{dt} \end{pmatrix} = \begin{pmatrix} 0.280\ \frac{m}{s}+2\cdot 0.0360\ \frac{m}{s^2}\cdot t \\3\cdot 0.0190\ \frac{m}{s^3}\cdot t^2 \end{pmatrix}\\\\ \text{b) } t = 5.01~s\\ v_x(5.01~s)&=&0.280\ \frac{m}{s}+2\cdot 0.0360\ \frac{m}{s^2}\cdot (5.01~s) = 0.64072~\frac{m}{s}\\ v_y(5.01~s)&=&3\cdot 0.0190\ \frac{m}{s^3}\cdot (5.01~s)^2= 1.4307057~\frac{m}{s}\\ v(5.01~s)&=&\sqrt{0.64072^2~(\frac{m}{s})^2+1.4307057^2~({\frac{m}{s}})^2} = 1.56762269645~\frac{m}{s}\\ \end{array}\)

 

The magnitude of the squirrel's velocity is 1.57 m/s

 

\(\begin{array}{rcl} \text{c) direction } &=& \arctan{ \left( \frac{v_y}{v_x} \right) } \\ &=& \arctan{ \left( \frac{1.4307057~\frac{m}{s}}{0.64072~\frac{m}{s}} \right) } \\ &=& \arctan{ \left(2.23296556998 \right) } \\ &=& 65.8754973481^{\circ} \end{array}\)

 

The direction (in degrees counterclockwise from +x-axis) of the squirrel's velocity is 65.9 degrees

 

laugh

heureka  Sep 8, 2015

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