(a) Answer v = 5.43 m/s Explanation Given information: Track: When rounding a curve, the acute angle θ that a runner’s body makes with the vertical is called the angle of incline. It is described by the equation tan θ = v 2 g r , 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: θ = 11 ∘ r = 15.5 m, g = 9.8 m s 2 tan θ = v 2 g r Plug in our values tan 11 ∘ = v 2 9.8 m s 2 15.5 m 0.1944 = v 2 151.9 m 2 s 2 Cross multiply 29.53 m 2 s 2 = v 2 Take the square root of each side v = 5.43 m/s (b) To determine To find: The runner’s velocity if the angle of incline is 13 ∘ . Answer v = 5.92 m/s Explanation Given information: Track: When rounding a curve, the acute angle θ that a runner’s body makes with the vertical is called the angle of incline. It is described by the equation tan θ = v 2 g r , 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: θ = 13 ∘ r = 15.5 m, g = 9.8 m s 2 tan θ = v 2 g r Plug in our values tan 13 ∘ = v 2 9.8 m s 2 15.5 m 0.2309 = v 2 151.9 m 2 s 2 Cross multiply 35.07 m 2 s 2 = v 2 Take the square root of each side v = 5.92 m/s (c) To determine To find: runner’s velocity if the angle of incline is 15 ∘ . Answer v = 6.38 m/s Explanation Given information: Track: When rounding a curve, the acute angle θ that a runner’s body makes with the vertical is called the angle of incline. It is described by the equation tan θ = v 2 g r , 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: θ = 15 ∘ r = 15.5 m, g = 9.8 m s 2 tan θ = v 2 g r Plug in our values tan 15 ∘ = v 2 9.8 m s 2 15.5 m 0.2679 = v 2 151.9 m 2 s 2 Cross multiply 40.69 m 2 s 2 = v 2 Take the square root of each side v = 6.38 m/s (d) To determine To find: whether a runner increase or decrease her velocity to increase his or her angle of incline. Answer She would increase it Explanation Given information: Track: When rounding a curve, the acute angle θ that a runner’s body makes with the vertical is called the angle of incline. It is described by the equation tan θ = v 2 g r , 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 θ = sin θ cos θ Thus, She would increase her velocity to increase his or her angle of incline.

Advanced Mathematical Concepts: Precalculus with Applications, Student Edition

1st Edition

ISBN: 9780078682278

Author: McGraw-Hill, Berchie Holliday

Publisher: Glencoe/McGraw-Hill

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Question

Chapter 5.2, Problem 29E

(a)

To determine

To find: runner’s velocity if the angle of incline is $11∘$ .

(a)

Expert Solution

Answer to Problem 29E

$v=5.43 m/s$

Explanation of Solution

Given information: Track: When rounding a curve, the acute angle $θ$ that a runner’s body makes with the vertical is called the angle of incline. It is described by the equation $tanθ=v2gr,$ 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:

$θ=11∘r=15.5m, g=9.8ms2tanθ=v2gr$

Plug in our values

$tan11∘=v29.8ms215.5m$

$0.1944=v2151.9m2s2$

Cross multiply

$29.53m2s2=v2$

Take the square root of each side

$v=5.43 m/s$

(b)

To determine

To find: The runner’s velocity if the angle of incline is $13∘$ .

(b)

Expert Solution

Answer to Problem 29E

$v=5.92 m/s$

Explanation of Solution

Calculation:

$θ=13∘r=15.5m, g=9.8ms2tanθ=v2gr$

Plug in our values

$tan13∘=v29.8ms215.5m$

$0.2309=v2151.9m2s2$

Cross multiply

$35.07m2s2=v2$

Take the square root of each side

$v=5.92 m/s$

(c)

To determine

To find: runner’s velocity if the angle of incline is $15∘$ .

(c)

Expert Solution

Answer to Problem 29E

$v=6.38 m/s$

Explanation of Solution

Calculation:

$θ=15∘r=15.5m, g=9.8ms2tanθ=v2gr$

Plug in our values

$tan15∘=v29.8ms215.5m$

$0.2679=v2151.9m2s2$

Cross multiply

$40.69m2s2=v2$

Take the square root of each side

$v=6.38 m/s$

(d)

To determine

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

(d)

Expert Solution

Answer to Problem 29E

She would increase it

Explanation of Solution

Calculation:

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

That is what we were asked to show

$tanθ=sinθcosθ$

Thus,

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

Chapter 5.2, Problem 28E

Chapter 5.2, Problem 30E

Chapter 5 Solutions

Advanced Mathematical Concepts: Precalculus with Applications, Student Edition

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