Chapter 5, Problem 49P

(a)

To determine

The speed of the car at $1/4$ of the circumference of the car traveled.

Answer to Problem 49P

The speed of the car is $17.7\text{m/s}$.

Explanation of Solution

Since the car moves with constant tangential acceleration with the circumference $C=2\pi r$.

Write the equation for the distance traveled by the car.

$\Delta x=\frac{C}{4}$ (I)

Here, $\Delta x$ is the distance traveled by the car and $C$ is the circumference of the car traveled.

Replace $2\pi r$ for $C$ in the equation (I).

$\begin{array}{c}\Delta x=\frac{2\pi r}{4}\\ =\frac{\pi r}{2}\end{array}$ (II)

Here, the radius of the circular path of the car traveled is $r$.

Write the equation of motion for the speed of the car.

$v_{\text{t}}^2=u^2+2a_{\text{t}}\Delta x$ (III)

Here, $v_{\text{t}}$ is the final tangential speed of the car, $u$ is the initial speed of the car, and $a_{\text{t}}$ is the tangential acceleration of the car traveled.

Conclusion:

Substitute equation (II) in equation (III) and $0\text{m/s}$ for $u$.

$\begin{array}{c}{v}_{\text{t}}^2={\left(0\text{m/s}\right)}^2+2a_{\text{t}}\left(\frac{\pi r}{2}\right)\\ =a_{\text{t}}\pi r\\ v_{\text{t}}=\sqrt{a_{\text{t}}\pi r}\end{array}$

Substitute $2.00{\text{m/s}}^2$ for $a_{\text{t}}$ and $50.0\text{m}$ for $r$ in above equation.

$\begin{array}{c}{v}_{\text{t}}=\sqrt{\left(2.00{\text{m/s}}^2\right)\left(3.14\right)\left(50.0\text{m}\right)}\\ =17.7\text{m/s}\end{array}$

Therefore, the speed of the car is $17.7\text{m/s}$.

(b)

To determine

The radial acceleration of the car at $1/4$ of the circumference of car traveled.

Answer to Problem 49P

The radial acceleration of the car is $6.28{\text{m/s}}^2$.

Explanation of Solution

Write the equation for radial acceleration of the car.

$a_{\text{r}}=\frac{v_{\text{t}}^2}{r}$                                                                                     (IV)

Here, $a_{\text{r}}$ is the radial acceleration of the car traveled at the circumference of $1/4$.

Conclusion:

Substitute $a\pi r$ for $v_{\text{t}}^2$ in equation (IV).

$\begin{array}{c}{a}_{\text{r}}=\frac{a_{\text{t}}\pi r}{r}\\ =a_{\text{t}}\pi \end{array}$ (V)

Substitute $2.00{\text{m/s}}^2$ for $a_{\text{t}}$ in equation (V).

$\begin{array}{c}{a}_{\text{r}}=\left(2.00{\text{m/s}}^2\right)\left(3.14\right)\\ =6.28{\text{m/s}}^2\end{array}$

Therefore, the radial acceleration of the car is $6.28{\text{m/s}}^2$.

(c)

To determine

The total acceleration of the car.

Answer to Problem 49P

The total acceleration of the car is $6.59{\text{m/s}}^2$ directed by an angle $17.7°$.

Explanation of Solution

The figure 1 shows the total acceleration exerted on the car.

Write the equation for total acceleration of the car.

$a=\sqrt{a_{\text{r}}^2+a_{\text{t}}^2}$                                                                                    (VI)

Here, $a$ is the total acceleration of the car traveled at the circumference of $1/4$.

Write the equation for the direction of the car travelled.

$\tan \theta =\frac{a_y}{a_x}$ (VII)

Conclusion:

Substitute $a_{\text{t}}\pi$ for $a_{\text{r}}$ in equation (VI).

$\begin{array}{c}a=\sqrt{a_{\text{t}}^2{\pi }^2+a_{\text{t}}^2}\\ =\sqrt{a_{\text{t}}^2\left({\pi }^2+1\right)}\\ =a_{\text{t}}\sqrt{\left({\pi }^2+1\right)}\end{array}$ (VIII)

Substitute $2.00{\text{m/s}}^2$ for $a_{\text{t}}$ in equation (VIII).

$\begin{array}{c}a=\left(2.00{\text{m/s}}^2\right)\sqrt{\left({3.14}^2+1\right)}\\ =6.59{\text{m/s}}^2\end{array}$

Therefore, the total acceleration of the car is $6.59{\text{m/s}}^2$.

Let us calculate the direction of the car travelled at the point of circumference $1/4$.

Consider $+y$ can be taken as north direction and $+x$ can be taken as east.

Since the acceleration for $+x$ is $a_x=a_{\text{t}}$, then acceleration for $+y$ is $a_y=-a_{\text{r}}$.

Rewrite the equation (VII) for $\theta$.

$\begin{array}{c}\theta ={\tan }^{-1}\frac{a_y}{a_x}\\ ={\tan }^{-1}\left(\frac{-a_{\text{r}}}{a_{\text{t}}}\right)\\ ={\tan }^{-1}\left(\frac{-\pi a_{\text{t}}}{a_{\text{t}}}\right)\\ ={\tan }^{-1}\left(-\pi \right)\end{array}$

Substitute $3.14$ for $\pi$ in above equation.

$\begin{array}{c}\theta ={\tan }^{-1}\left(-3.14\right)\\ =-72.3°\\ \varphi =\left(90°-72.3°\right)\\ =17.7°\end{array}$

Therefore, the total acceleration of the car is $6.59{\text{m/s}}^2$ directed along the east of south direction at an angle $17.7°$.