Chapter 5.2, Problem 29E

(a)

To determine

To find: runner’s velocity if the angle of incline is ${11}^{\circ }$ .

Answer to Problem 29E

  $v=5.43\text{ m/s}$

Explanation of Solution

Given information: Track: When rounding a curve, the acute angle $\theta$ that a runner’s body makes with the vertical is called the angle of incline. It is described by the equation $\tan \theta =\frac{v^2}{gr},$ where v is the velocity of the runner, g is the acceleration due to gravity and r is the radius of the track. The acceleration due to gravity is a constant 9.8 meters per second squared. Suppose the radius of the track is 15.5 meters.

Calculation:

  $\begin{array}{l}\theta ={11}^{\circ }\\ r=15.5\text{m, }g=\frac{9.8\text{m}}{{\text{s}}^2}\\ \tan \theta =\frac{v^2}{gr}\end{array}$

Plug in our values

  $\tan {11}^{\circ }=\frac{v^2}{\frac{9.8\text{m}}{{\text{s}}^2}15.5\text{m}}$

  $0.1944=\frac{v^2}{151.9\frac{{\text{m}}^2}{{\text{s}}^2}}$

Cross multiply

  $29.53\frac{{\text{m}}^2}{{\text{s}}^2}=v^2$

Take the square root of each side

  $v=5.43\text{ m/s}$

(b)

To determine

To find: The runner’s velocity if the angle of incline is ${13}^{\circ }$ .

Answer to Problem 29E

  $v=5.92\text{ m/s}$

Explanation of Solution

Given information: Track: When rounding a curve, the acute angle $\theta$ that a runner’s body makes with the vertical is called the angle of incline. It is described by the equation $\tan \theta =\frac{v^2}{gr},$ where v is the velocity of the runner, g is the acceleration due to gravity and r is the radius of the track. The acceleration due to gravity is a constant 9.8 meters per second squared. Suppose the radius of the track is 15.5 meters.

Calculation:

  $\begin{array}{l}\theta ={13}^{\circ }\\ r=15.5\text{m, }g=\frac{9.8\text{m}}{{\text{s}}^2}\\ \tan \theta =\frac{v^2}{gr}\end{array}$

Plug in our values

  $\tan {13}^{\circ }=\frac{v^2}{\frac{9.8\text{m}}{{\text{s}}^2}15.5\text{m}}$

  $0.2309=\frac{v^2}{151.9\frac{{\text{m}}^2}{{\text{s}}^2}}$

Cross multiply

  $35.07\frac{{\text{m}}^2}{{\text{s}}^2}=v^2$

Take the square root of each side

  $v=5.92\text{ m/s}$

(c)

To determine

To find: runner’s velocity if the angle of incline is ${15}^{\circ }$ .

Answer to Problem 29E

  $v=6.38\text{ m/s}$

Explanation of Solution

Given information: Track: When rounding a curve, the acute angle $\theta$ that a runner’s body makes with the vertical is called the angle of incline. It is described by the equation $\tan \theta =\frac{v^2}{gr},$ where v is the velocity of the runner, g is the acceleration due to gravity and r is the radius of the track. The acceleration due to gravity is a constant 9.8 meters per second squared. Suppose the radius of the track is 15.5 meters.

Calculation:

  $\begin{array}{l}\theta ={15}^{\circ }\\ r=15.5\text{m, }g=\frac{9.8\text{m}}{{\text{s}}^2}\\ \tan \theta =\frac{v^2}{gr}\end{array}$

Plug in our values

  $\tan {15}^{\circ }=\frac{v^2}{\frac{9.8\text{m}}{{\text{s}}^2}15.5\text{m}}$

  $0.2679=\frac{v^2}{151.9\frac{{\text{m}}^2}{{\text{s}}^2}}$

Cross multiply

  $40.69\frac{{\text{m}}^2}{{\text{s}}^2}=v^2$

Take the square root of each side

  $v=6.38\text{ m/s}$

(d)

To determine

To find: whether a runner increase or decrease her velocity to increase his or her angle of incline.

Answer to Problem 29E

She would increase it

Explanation of Solution

Given information: Track: When rounding a curve, the acute angle $\theta$ that a runner’s body makes with the vertical is called the angle of incline. It is described by the equation $\tan \theta =\frac{v^2}{gr},$ where v is the velocity of the runner, g is the acceleration due to gravity and r is the radius of the track. The acceleration due to gravity is a constant 9.8 meters per second squared. Suppose the radius of the track is 15.5 meters.

Calculation:

For the three previous problems, the velocity increase made the angle increase.

That is what we were asked to show

  $\tan \theta =\frac{\sin \theta }{\cos \theta }$

Thus,

She would increase her velocity to increase his or her angle of incline.