IDENTIFY and SET UP:
This problem involves motion in two dimensions, so we must use the vector equations obtained in this section. Figure 3.5 shows the rover’s path (dashed line). We’ll use Eq. (3.1) (\vec{r}={x} \hat{\imath}+{y} \hat{j}+{z} \hat{k}) for position \overrightarrow{\boldsymbol{r}}, the expression \Delta \vec{r}=\vec{r}_{2}-\vec{r}_{1} for displacement, Eq. (3.2) (\overrightarrow{\boldsymbol{v}}_{\mathrm{av}}=\frac{\Delta \overrightarrow{\boldsymbol{r}}}{\Delta t}=\frac{\overrightarrow{\boldsymbol{r}}_{2}-\overrightarrow{\boldsymbol{r}}_{1}}{t_{2}-t_{1}}) for average velocity, and Eqs. (3.5) (\overrightarrow{\boldsymbol{v}}=\frac{d \overrightarrow{\boldsymbol{r}}}{d t}=\frac{d x}{d t} \hat{\imath}+\frac{d y}{d t} \hat{\jmath}+\frac{d z}{d t} \hat{\boldsymbol{k}}), (3.6) (|\overrightarrow{\boldsymbol{v}}|=v=\sqrt{v_{x}^{2}+v_{y}^{2}+v_{z}^{2}}), and (3.7) (\tan \alpha=\frac{v_{y}}{v_{x}}) for instantaneous velocity and its magnitude and direction.
EXECUTE:
(a) At t = 2.0 s the rover’s coordinates are
x=2.0 \mathrm{~m}-\left(0.25 \mathrm{~m} / \mathrm{s}^{2}\right)(2.0 \mathrm{~s})^{2}=1.0 \mathrm{~m}
y=(1.0 \mathrm{~m} / \mathrm{s})(2.0 \mathrm{~s})+\left(0.025 \mathrm{~m} / \mathrm{s}^{3}\right)(2.0 \mathrm{~s})^{3}=2.2 \mathrm{~m}
The rover’s distance from the origin at this time is
r=\sqrt{x^{2}+y^{2}}=\sqrt{(1.0 \mathrm{~m})^{2}+(2.2 \mathrm{~m})^{2}}=2.4 \mathrm{~m}
(b) To find the displacement and average velocity over the given time interval, we first express the position vector \overrightarrow{\boldsymbol{r}} as a function of time t. From Eq. (3.1) (\vec{r}={x} \hat{\imath}+{y} \hat{j}+{z} \hat{k}) for position \overrightarrow{\boldsymbol{r}} this is
\overrightarrow{\boldsymbol{r}}=x \hat{\boldsymbol{i}}+y \hat{\jmath}=\left[2.0 \mathrm{~m}-\left(0.25 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}\right] \hat{\imath}+\left[(1.0 \mathrm{~m} / \mathrm{s}) t+\left(0.025 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3}\right] \hat{\jmath}
At t = 0.0 s the position vector \overrightarrow{\boldsymbol{r}}_{0} is
\vec{r}_{0}=(2.0 \mathrm{~m}) \hat{\imath}+(0.0 \mathrm{~m}) \hat{\jmath}
From part (a), the position vector \overrightarrow{\boldsymbol{r}}_{2} at t = 2.0 s is
\overrightarrow{\boldsymbol{r}}_{2}=(1.0 \mathrm{~m}) \hat{\imath}+(2.2 \mathrm{~m}) \hat{\jmath}
The displacement from t = 0.0 s to t = 2.0 s is therefore
\Delta \overrightarrow{\boldsymbol{r}}=\overrightarrow{\boldsymbol{r}}_{2}-\overrightarrow{\boldsymbol{r}}_{0}=(1.0 \mathrm{~m}) \hat{\imath}+(2.2 \mathrm{~m}) \hat{\jmath}-(2.0 \mathrm{~m}) \hat{\imath}
=(-1.0 \mathrm{~m}) \hat{\imath}+(2.2 \mathrm{~m}) \hat{\jmath}
During this interval the rover moves 1.0 m in the negative x-direction and 2.2 m in the positive y-direction. From Eq. (3.2) (\overrightarrow{\boldsymbol{v}}_{\mathrm{av}}=\frac{\Delta \overrightarrow{\boldsymbol{r}}}{\Delta t}=\frac{\overrightarrow{\boldsymbol{r}}_{2}-\overrightarrow{\boldsymbol{r}}_{1}}{t_{2}-t_{1}}), the average velocity over this interval is the displacement divided by the elapsed time:
\vec{v}_{\mathrm{av}}=\frac{\Delta \vec{r}}{\Delta t}=\frac{(-1.0 \mathrm{~m}) \hat{\imath}+(2.2 \mathrm{~m}) \hat{\jmath}}{2.0 \mathrm{~s}-0.0 \mathrm{~s}}=(-0.50 \mathrm{~m} / \mathrm{s}) \hat{\imath}+(1.1 \mathrm{~m} / \mathrm{s}) \hat{\jmath}
The components of this average velocity are v_{av-x} = -0.50 m/s and v_{av-x} = 1.1 m/s.
(c) From Eq. (3.4) (v_{x}=\frac{d x}{d t} \quad v_{y}=\frac{d y}{d t} \quad v_{z}=\frac{d z}{d t}) the components of instantaneous velocity are the time derivatives of the coordinates:
v_{x}=\frac{d x}{d t}=\left(-0.25 \mathrm{~m} / \mathrm{s}^{2}\right)(2 t)
v_{y}=\frac{d y}{d t}=1.0 \mathrm{~m} / \mathrm{s}+\left(0.025 \mathrm{~m} / \mathrm{s}^{3}\right)\left(3 t^{2}\right)
Hence the instantaneous velocity vector is
\overrightarrow{\boldsymbol{v}}=v_{x} \hat{\imath}+v_{y} \hat{\jmath}=\left(-0.50 \mathrm{~m} / \mathrm{s}^{2}\right) t \hat{\imath}+\left[1.0 \mathrm{~m} / \mathrm{s}+\left(0.075 \mathrm{~m} / \mathrm{s}^{3}\right) t^{2}\right] \hat{\jmath}
At t = 2.0 s the velocity vector \overrightarrow{\boldsymbol{r}}_{2} has components
v_{2 x}=\left(-0.50 \mathrm{~m} / \mathrm{s}^{2}\right)(2.0 \mathrm{~s})=-1.0 \mathrm{~m} / \mathrm{s}
v_{2 y}=1.0 \mathrm{~m} / \mathrm{s}+\left(0.075 \mathrm{~m} / \mathrm{s}^{3}\right)(2.0 \mathrm{~s})^{2}=1.3 \mathrm{~m} / \mathrm{s}
The magnitude of the instantaneous velocity (that is, the speed) at t = 2.0 s is
v_{2}=\sqrt{v_{2 x}^{2}+v_{2 y}^{2}}=\sqrt{(-1.0 \mathrm{~m} / \mathrm{s})^{2}+(1.3 \mathrm{~m} / \mathrm{s})^{2}}= 1.6 m/s
Figure 3.5 shows the direction of velocity vector \overrightarrow{\boldsymbol{r}}_{2}, which is at an
angle a between 90° and 180° with respect to the positive x-axis.
From Eq. (3.7) (\tan \alpha=\frac{v_{y}}{v_{x}}) we have
\arctan \frac{v_{y}}{v_{x}}=\arctan \frac{1.3 \mathrm{~m} / \mathrm{s}}{-1.0 \mathrm{~m} / \mathrm{s}}=-52^{\circ}
This is off by 180°; the correct value is a = 180° – 52° = 128°, or 38° west of north.
EVALUATE: Compare the components of average velocity from part (b) for the interval from t = 0.0 s to t = 2.0 s (v_{av-x} = -0.50 m/s, v_{av-x} = 1.1 m/s) with the components of instantaneous velocity at t = 2.0 s from part (c) (v_{2x} = -1.0 m/s, v_{2y} = 1.3 m/s). Just as in one dimension, the average velocity vector \overrightarrow{\boldsymbol{v}}_{\mathrm{av}} over an interval is in general not equal to the instantaneous
velocity \overrightarrow{\boldsymbol{v}} at the end of the interval (see Example 2.1).
Figure 3.5 shows the position vectors \overrightarrow{\boldsymbol{r}} and instantaneous velocity vectors \overrightarrow{\boldsymbol{v}} at t = 0.0 s, 1.0 s, and 2.0 s. (Calculate these quantities for t = 0.0 s and t = 1.0 s.) Notice that \overrightarrow{\boldsymbol{v}} is tangent to the path at every point. The magnitude of \overrightarrow{\boldsymbol{v}} increases as the rover moves, which means that its speed is increasing.
