Chapter 5, Problem 49P

(a)

To determine

The speed of the car after travelled one fourth of the circumference of the circular path.

Answer to Problem 49P

The speed of the car after travelled one fourth of the circumference of the circular path is $\underset{\_}{17.7\text{m}/\text{s}}$.

Explanation of Solution

Write the expression for the circumference of the circle.

    $C=2\pi r$        (I)

Here, $C$ is the circumference of the circle, $r$ is the radius of the circle.

Write the expression for the kinematic equation for the velocity.

    $v^2=u^2+2a_{T}s$        (II)

Here, $v$ is the final velocity, $a_T$ is the acceleration, $s$ is the distance travelled by the car.

The initial velocity of the car is zero and the distance travelled is one fourth of the circumference of the circular path.

Use equation (I) and (II) to write the expression of velocity for the one fourth of the circumference of the circular path travelled by the car.

    $v=\sqrt{2a_T\left(\frac{C}{4}\right)}$        (III)

Use equation (I) in (III) to solve for $v$.

    $\begin{array}{c}v=\sqrt{2a_T\left(\frac{2\pi r}{4}\right)}\\ =\sqrt{a_T\pi r}\end{array}$        (IV)

Conclusion:

Substitute $2.00\text{m}/{\text{s}}^{\text{2}}$ for $a_T$, $50.0\text{m}$ for $r$ in equation (IV) to find $v$.

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

Therefore, the speed of the car after travelled one fourth of the circumference of the circular path is $\underset{\_}{17.7\text{m}/\text{s}}$.

(b)

To determine

The radial acceleration of the car.

Answer to Problem 49P

The radial acceleration of the car is $\underset{\_}{6.26\text{m}/{\text{s}}^2}$.

Explanation of Solution

Write the expression for the radial acceleration.

    $a_r=\frac{v^2}{r}$        (V)

Here, $a_r$ is the radial acceleration.

Conclusion:

Substitute $17.7\text{m}/\text{s}$ for $v$, $50.0\text{m}$ for $r$ in equation (V) to find $a_r$.

    $\begin{array}{c}{a}_r=\frac{{\left(17.7\text{m}/{\text{s}}^2\right)}^2}{50.0\text{m}}\\ =6.26\text{m}/{\text{s}}^2\end{array}$

Therefore, the radial acceleration of the car is $\underset{\_}{6.26\text{m}/{\text{s}}^2}$.

(c)

To determine

The total acceleration of the car.

Answer to Problem 49P

The total acceleration of the car is $\underset{\_}{6.57\text{m}/{\text{s}}^2}$.

Explanation of Solution

Write the expression for the total acceleration.

    $\overrightarrow{a}=a_T\left(-\widehat{i}\right)+a_r\left(-\widehat{j}\right)$        (VI)

Here, $a$ is the total acceleration.

Write the expression for the magnitude of $\overrightarrow{a}$.

    $a=\sqrt{a_T^2+a_r^2}$        (VII)

Conclusion:

Substitute $2.00\text{m}/{\text{s}}^2$ for $a_T$, $6.26\text{m}/{\text{s}}^2$ for $a_r$ in equation (VII) to find $a$.

    $\begin{array}{c}a=\sqrt{{\left(2.00\text{m}/{\text{s}}^2\right)}^2+{\left(6.26\text{m}/{\text{s}}^2\right)}^2}\\ =6.57\text{m}/{\text{s}}^2\end{array}$

Therefore, the total acceleration of the car is $\underset{\_}{6.57\text{m}/{\text{s}}^2}$.