Chapter 11, Problem 10AP

A speed skater increases her speed from 10 m/s to 12.5 m/s over a period of 3 s while coming out of a curve of 20-m radius. What are the magnitudes of her radial, tangential, and total accelerations as she leaves the curve? (Remember that a_r and a_t are the vector components of total acceleration.) (Answer: a_r = 7.81 m/s²; a_t = 0.83 m/s²; a = 7.85 m/s²)

Summary Introduction

To determine: The radial acceleration of the skater.

Answer to Problem 10AP

The magnitude of radial acceleration is $7.81\text{m}/{\text{s}}^2$.

Explanation of Solution

Calculation:

Express the relation for radial acceleration.

$a_r=\frac{v^2}{r}$

Here, $a_r$ is the radial acceleration, $v$ is the linear velocity, and $r$ is the radius of the curve.

Substitute $12.5\text{m}/\text{s}$ for $v$ and $20\text{m}$ for $r$ to find the radial acceleration.

$\begin{array}{c}{a}_r=\frac{{\left(12.5\text{m}/\text{s}\right)}^2}{20\text{m}}\\ =\frac{156.25{\text{m}}^2/{\text{s}}^2}{20\text{m}}\\ =7.8125\text{m}/{\text{s}}^2\\ \approx 7.81\text{m}/{\text{s}}^2\end{array}$

Therefore, the magnitude of radial acceleration is $7.81\text{m}/{\text{s}}^2$.

Conclusion

Therefore, the magnitude of radial acceleration is $7.81\text{m}/{\text{s}}^2$.

Summary Introduction

To determine: The tangential acceleration of the skater.

Answer to Problem 10AP

The magnitude of tangential acceleration is $0.83\text{m}/{\text{s}}^2$.

Explanation of Solution

Calculation:

Express the relation for tangential acceleration.

$a_t=\left(\frac{\text{Final}\text{velocity}-\text{initial}\text{velocity}}{\text{time}}\right)$

Here, $a_t$ is the tangential acceleration.

Substitute $12.5\text{m}/\text{s}$ for final velocity, $10\text{m}/\text{s}$ for initial velocity, and $3\text{s}$ for time to find the tangential acceleration.

$\begin{array}{c}{a}_t=\left(\frac{12.5\text{m}/\text{s}-10\text{m}/\text{s}}{\text{3}\text{s}}\right)\\ =\left(\frac{2.5\text{m}/\text{s}}{3\text{s}}\right)\\ =0.8333\text{m}/{\text{s}}^2\\ \approx 0.83\text{m}/{\text{s}}^2\end{array}$

Therefore, the magnitude of tangential acceleration is $0.83\text{m}/{\text{s}}^2$.

Conclusion

Therefore, the magnitude of tangential acceleration is $0.83\text{m}/{\text{s}}^2$.

Summary Introduction

To determine: The total acceleration of the skater.

Answer to Problem 10AP

The magnitude of total acceleration is $7.85\text{m}/{\text{s}}^2$.

Explanation of Solution

Calculation:

Express the relation for total acceleration.

$a=\sqrt{a_r^2+a_t^2}$

Here, $a$ is the total acceleration.

Substitute  $7.81\text{m}/{\text{s}}^2$ for $a_r$ and $0.83\text{m}/{\text{s}}^2$ for $a_t$ to find the acceleration.

$\begin{array}{c}a=\sqrt{{\left(7.81\text{m}/{\text{s}}^2\right)}^2+{\left(0.83\text{m}/{\text{s}}^2\right)}^2}\\ =7.85\text{m}/{\text{s}}^2\end{array}$

Therefore, the magnitude of total acceleration is $7.85\text{m}/{\text{s}}^2$.

Conclusion

Therefore, the magnitude of total acceleration is $7.85\text{m}/{\text{s}}^2$.