Chapter 3, Problem 47P

(a)

To determine

The average velocity of the car in the whole trip.

Answer to Problem 47P

The average velocity of the car in the whole trip is $\underset{\_}{3.33\text{m/s, 45}°\text{ north of east}}$.

Explanation of Solution

Given that the radius of the circular path is $20.0\text{m}$, and the time taken for covering three quarters of the circle is $8.50\text{s}$ with constant speed.

The vector diagram indicating the initial velocity ${\overrightarrow{v}}_{\text{i}}$, and the final velocity ${\overrightarrow{v}}_{\text{f}}$ is given in Figure 1. The center of the circular path is considered as the origin.

The $+y$ direction represents the north, and $+x$ direction represents the east.

Write the expression for the displacement vector of the car.

$\Delta \overrightarrow{r}={\overrightarrow{r}}_{\text{f}}-{\overrightarrow{r}}_{\text{i}}$ (I)

Here, $\Delta \overrightarrow{r}$ is the displacement vector, ${\overrightarrow{r}}_{\text{f}}$ is the final position vector, and ${\overrightarrow{r}}_{\text{i}}$ is the initial position vector.

The initial position vector of the car is $-20.0\text{m}\widehat{x}$, and the final position vector is $-20.0\text{m}\widehat{y}$.

Write the expression for the magnitude of the displacement vector.

$\left|\Delta \overrightarrow{r}\right|=\sqrt{r_x^2+r_y^2}$ (II)

Here, $r_x$ is the of $x$-component of $\Delta \overrightarrow{r}$, and $r_y$ is the of $y$-component of $\Delta \overrightarrow{r}$.

The average velocity will be in the same direction of the displacement vector.

Write the expression for the angle $\theta$ that the displacement vector (as well as the average velocity vector) makes with the $+x$-direction (east).

$\theta ={\tan }^{-1}\left(\frac{r_y}{r_x}\right)$ (III)

Write the expression for the magnitude of the average velocity.

$\left|{\overrightarrow{v}}_{\text{av}}\right|=\frac{\left|\Delta \overrightarrow{r}\right|}{\Delta t}$ (IV)

Here, ${\overrightarrow{v}}_{\text{av}}$ is the average velocity vector, and $\Delta t$ is the time duration.

Conclusion:

Substitute $20.0\text{m}\widehat{x}$ for ${\overrightarrow{r}}_{\text{f}}$, and $-20.0\text{m}\widehat{y}$ for ${\overrightarrow{r}}_{\text{i}}$ in equation (I) to find $\Delta \overrightarrow{r}$.

$\begin{array}{c}\Delta \overrightarrow{r}=20.0\text{m}\widehat{x}-\left(-20.0\text{m}\widehat{y}\right)\\ =20.0\text{m}\widehat{x}+20.0\text{m}\widehat{y}\end{array}$

Substitute $20.0\text{m}$ for $r_x$, and $20.0\text{m}$ for $r_y$ in equation (II) to find $\left|\Delta \overrightarrow{r}\right|$.

$\begin{array}{c}\left|\Delta \overrightarrow{r}\right|=\sqrt{{\left(20.0\text{m}\right)}^2+{\left(20.0\text{m}\right)}^2}\\ =28.3\text{m}\end{array}$

Substitute $20.0\text{m}$ for $r_x$, and $20.0\text{m}$ for $r_y$ in equation (III) to find $\theta$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{20.0\text{m}}{20.0\text{m}}\right)\\ =+45°\\ \Rightarrow 45°\text{north of east}\end{array}$

Substitute $28.3\text{m}$ for $\left|\Delta \overrightarrow{r}\right|$, and $8.50\text{s}$ for $\Delta t$ in equation (IV) to find $\left|{\overrightarrow{v}}_{\text{av}}\right|$.

$\begin{array}{c}\left|{\overrightarrow{v}}_{\text{av}}\right|=\frac{28.3\text{m}}{8.50\text{s}}\\ =3.33\text{m/s}\end{array}$

Combine the magnitude and direction of the average velocity vector.

${\overrightarrow{v}}_{\text{av}}=3.33\text{m/s, 45}°\text{ north of east}$.

Therefore, the average velocity of the car in the whole trip is $\underset{\_}{3.33\text{m/s, 45}°\text{ north of east}}$.

(b)

To determine

The average acceleration of the car in the whole trip.

Answer to Problem 47P

The average acceleration of the car in the whole trip is $\underset{\_}{1.84{\text{m/s}}^2,\text{ 45}°\text{ south of east}}$.

Explanation of Solution

It is obtained that the average velocity of the car in the whole trip is $3.33\text{m/s, 45}°\text{ north of east}$.

Write the expression for the average acceleration.

${\overrightarrow{a}}_{\text{av}}=\frac{{\overrightarrow{v}}_{\text{f}}-{\overrightarrow{v}}_{\text{i}}}{\Delta t}$ (V)

Here, ${\overrightarrow{a}}_{\text{av}}$ is the average acceleration, ${\overrightarrow{v}}_{\text{f}}$ is the final velocity, and ${\overrightarrow{v}}_{\text{i}}$ is the initial velocity.

Since the car covers three quarters of the circle, the distance travelled is obtained as,

$d=\left(3/4\right)2\pi r$ (VI)

Here, $d$ is the distance covered, and $r$ is the radius of the circular track.

The initial velocity is directed west and the final velocity is directed south. The expression for the initial and final velocity is obtained as,

${\overrightarrow{v}}_{\text{i}}=-\frac{d}{\Delta t}\widehat{x}$

${\overrightarrow{v}}_{\text{f}}=-\frac{d}{\Delta t}\widehat{y}$

Use the expression for ${\overrightarrow{v}}_{\text{i}}$ and ${\overrightarrow{v}}_{\text{f}}$ in equation (V).

$\begin{array}{c}{\overrightarrow{a}}_{\text{av}}=\frac{\left(-\frac{d}{\Delta t}\widehat{y}\right)-\left(-\frac{d}{\Delta t}\widehat{x}\right)}{\Delta t}\\ =\frac{\frac{d}{\Delta t}\left(\widehat{x}-\widehat{y}\right)}{\Delta t}\\ =\frac{d}{{\left(\Delta t\right)}^2}\left(\widehat{x}-\widehat{y}\right)\end{array}$ (VII)

Use equation (VI) in (VII).

$\begin{array}{c}{\overrightarrow{a}}_{\text{av}}=\frac{\left(3/4\right)2\pi r}{{\left(\Delta t\right)}^2}\left(\widehat{x}-\widehat{y}\right)\\ =\frac{3\pi r}{2{\left(\Delta t\right)}^2}\left(\widehat{x}-\widehat{y}\right)\end{array}$ (VIII)

Write the expression for the magnitude of average acceleration.

$\left|{\overrightarrow{a}}_{\text{av}}\right|=\sqrt{a_x^2+a_y^2}$ (IX)

Here, $a_x$ is the $x$-component of ${\overrightarrow{a}}_{\text{av}}$, and $a_y$ is the $y$-component of ${\overrightarrow{a}}_{\text{av}}$.

Conclusion:

From equation (VIII) it is clear that the average acceleration vector has $x$-component $\frac{3\pi r}{2{\left(\Delta t\right)}^2}$, and $y$-component $\left(-\frac{3\pi r}{2{\left(\Delta t\right)}^2}\right)$.

Substitute $\left(3\pi r/2{\left(\Delta t\right)}^2\right)$ for $a_x$, and $\left(-3\pi r/2{\left(\Delta t\right)}^2\right)$ for $a_y$ in equation (IX) to obtain the expression for $\left|{\overrightarrow{a}}_{\text{av}}\right|$.

$\begin{array}{c}\left|{\overrightarrow{a}}_{\text{av}}\right|=\sqrt{{\left(\frac{3\pi r}{2{\left(\Delta t\right)}^2}\right)}^2+{\left(-\frac{3\pi r}{2{\left(\Delta t\right)}^2}\right)}^2}\\ =\frac{3\pi r}{\sqrt{2}{\left(\Delta t\right)}^2}\end{array}$ (X)

Write the expression for the angle $\theta$ that the average acceleration vector makes with the $+x$-direction (east).

$\theta ={\tan }^{-1}\left(\frac{a_y}{a_x}\right)$ (XI)

Substitute $20.0\text{m}$ for $r$, and $8.50\text{s}$ for $\Delta t$ in equation (X) to find $\left|{\overrightarrow{a}}_{\text{av}}\right|$.

$\begin{array}{c}\left|{\overrightarrow{a}}_{\text{av}}\right|=\frac{3\pi \left(20.0\text{m}\right)}{\sqrt{2}{\left(8.50\text{s}\right)}^2}\\ =1.84{\text{m/s}}^2\end{array}$

Substitute $1.84{\text{m/s}}^2$ for $a_x$,and $-1.84{\text{m/s}}^2$ for $a_y$ in equation (XI) to find $\theta$

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-1.84{\text{m/s}}^2}{1.84{\text{m/s}}^2}\right)\\ =-45°\\ \Rightarrow 45°\text{south of east}\end{array}$

Combine the magnitude and direction of the average acceleration vector.

${\overrightarrow{a}}_{\text{av}}=1.84{\text{m/s}}^2\text{, 45}°\text{ south of east}$.

Therefore, the average acceleration of the car in the whole trip is $\underset{\_}{1.84{\text{m/s}}^2,\text{ 45}°\text{ south of east}}$.

(c)

To determine

The reason for which a car moving at constant speed in a circular path has a nonzero average acceleration.

Answer to Problem 47P

The change in direction of velocity requires an acceleration, which leads to the presence of a nonzero acceleration to a car moving in a circular path with constant speed.

Explanation of Solution

The direction of instantaneous velocity of the car moving in a circular path, changes in each instant. The direction of velocity at a point on the circular path will be along the tangent drawn at each of those points. Even though the magnitude of velocity (which is the speed) of motion is constant, due to the direction change of the velocity vector, the car gains an acceleration.

Therefore, the change in direction of velocity requires an acceleration, which leads to the presence of a nonzero acceleration to a car moving in a circular path with constant speed.