Answer
a.) $v_{avg}=7.1cm/s$, $\theta=45^{\circ}$
b.) $v=5cm/s$, $\theta=90^{\circ}$ - $v_{avg}=7.1cm/s$, $\theta=45^{\circ}$ - $v_{avg}=11.1cm/s$, $\theta=27^{\circ}$
c.) (graph)
Work Step by Step
a.) In obtaining the average velocity using the given position vector, use the formula $$\vec{v_{avg}}=\frac{\vec{r}_{2}-\vec{r}_{1}}{t_{2}-t_{1}}$$ where $\vec{r}_2$ is the position vector at $t=2.0 s$ and $\vec{r}_1$ is the position vector at $t=0 s$.
Substituting the change in time into the position vector,
$$\vec{r}_{2}=[4.0cm+(2.5cm/s^{2})(2.0s)^{2}]\hat{i}+5.0cm/s(2.0s)\hat{j}
=[4.0cm+10cm]\hat{i}+10cm\hat{j}=14cm\hat{i}+10cm\hat{j}$$
$$\vec{r}_{1}=[4.0cm+(2.5cm/s^{2})(0s)^{2}]\hat{i}+5.0cm/s(0s)\hat{j}
=[4.0cm+0cm]\hat{i}+0cm\hat{j}=4cm\hat{i}$$
And then substituting the position vectors into the formula for average velocity,
$$\vec{v_{avg}}=\frac{[14cm\hat{i}+10cm\hat{j}]-4cm\hat{i}}{2.0s-0s}
=\frac{10cm\hat{i}+10cm\hat{j}}{2.0s}=10cm\hat{i}+10cm\hat{j}$$
Where the magnitude can be calculated using the following
$$v_{avg}=\sqrt{(10cm/s)^{2}+(10cm/s)^{2}}=7.1cm/s$$
and the direction using tangent,
$$\theta=\arctan\frac{10}{10}=45^{\circ}$$
b.) In the case of instantaneous velocity, the velocity formula changes into
$$\vec{v}=v_{x}\hat{i}+v_{y}\hat{j}$$
Since a velocity vector can be defined as the first derivative of the position vector, evaluating the first derivative of the position vector
$$\vec{v}=5cm/s(t)\hat{i}+5cm/s\hat{j}$$
Then finally substituting the given times,
at $t=0s$,
$$\vec{v}=5cm/s(0s)\hat{i}+5cm/s\hat{j}=5cm/s\hat{j}$$
$$v=\sqrt{(5cm/s)^{2}}=5cm/s$$
Since the vector is positive and $\hat{j}$, the direction is $90^{\circ}$.
at $t=1s$,
$$\vec{v}=5cm/s\hat{i}+5cm/s\hat{j}$$
This is similar to the answer in a. where $v=7.1cm/s$
and the direction $\theta=45^{\circ}$.
at $t=2s$,
$$\vec{v}=5cm/s(2s)\hat{i}+5cm/s\hat{j}=10cm/s\hat{i}+5cm/s\hat{j}$$
$$v=\sqrt{(10cm/s)^{2}+(5cm/s)^{2}}=11.1cm/s$$
Determining the direction using tangent,
$$\theta=\arctan\frac{5}{10}=27^{\circ}$$
c. Using an online graphing calculator, input the position vector and substitute time points,
