Question

The position of a squirrel running in a park is given by $\vec{r} = \left( (0.280 \text{ m/s})t \hat{i} + (0.0360 \text{ m/s}^2)t^2 \hat{j} + (0.0190 \text{ m/s}^3)t^3 \hat{k} \right)$. Correct Part D At $t = 5.60 \text{ s}$, what is the magnitude of the squirrel's velocity? Express your answer with the appropriate units. Value Units Submit Request Answer Part E At $t = 5.60 \text{ s}$, what is the direction (in degrees counterclockwise from $+x$-axis) of the squirrel's velocity? Express your answer in degrees. $\theta = $

          The position of a squirrel running in a park is given by
$\vec{r} = \left( (0.280 \text{ m/s})t \hat{i} + (0.0360 \text{ m/s}^2)t^2 \hat{j} + (0.0190 \text{ m/s}^3)t^3 \hat{k} \right)$.
Correct
Part D
At $t = 5.60 \text{ s}$, what is the magnitude of the squirrel's velocity?
Express your answer with the appropriate units.
Value
Units
Submit
Request Answer
Part E
At $t = 5.60 \text{ s}$, what is the direction (in degrees counterclockwise from $+x$-axis) of the squirrel's velocity?
Express your answer in degrees.
$\theta = $

The position of a squirrel running in a park is given by
r⃗ = ( (0.280 m/s)t î + (0.0360 m/s^2)t^2 ĵ + (0.0190 m/s^3)t^3 k̂).
Correct
Part D
At t = 5.60 s, what is the magnitude of the squirrel's velocity?
Express your answer with the appropriate units.
Value
Units
Submit
Request Answer
Part E
At t = 5.60 s, what is the direction (in degrees counterclockwise from +x-axis) of the squirrel's velocity?
Express your answer in degrees.
θ =

Added by Jordan R.

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