Question

A runner sprints around a circular track of radius 130 m at a constant speed of 7 m/s. The runner's friend is standing at a distance 260 m from the center of the track. How fast is the distance between the friends changing when the distance between them is 260 m? (Round your answer to two decimal places.)\

Answer 1

Answer: 6.78 m/sec

Explanation:

Let the track be centered at origin, with radius = 130

Let θ = angle formed by line from centre to runner with positive x-axis

Let S = distance along circular track where runner is located

and where S = 0 is at position (130,0)

Runner sprints at constant speed of 7m/sec . . . . ds/dt = 7

Formula for arc length:

s = r θ

s = 130 θ

d/dt = 130 dθ/dt

7 = 130 dθ/dt

dθ/dt = 7/130

Now,

Let (x,y) be runners position

x = r cos(θ) = 130 cos(θ)

y = r sin(θ) = 130 sin(θ)

x² + y² = r² = 16,900

Let runner's friend be at position (260, 0) (angle of 0 degrees with positive x-axis)

Let D = distance from runner to runner's friend

D² = (260 - x)² + y²

D² = 67,600 - 520x + x² + y²

D² = 67,600 - 520x + 16,900

D² = 84,500 - 520(130 cos(θ))

D² = 84,500 - 67,600 cos(θ)

Differentiating both sides with respect to t, we get

2D dD/dt = 67,600 sin(θ) dθ/dt

Find dD/dt when D = 260

First, we find θ when D = 260 using formula

D² = 84,500 - 67,600 cos(θ)

67,600 = 84,500 - 67,600 cos(θ)

67,600 cos(θ) = 16,900

cos(θ) = 1/4

sin(θ) = √(1 - (1/4)²) = √(15/16) = √15/4

Now we can calculate dD/dt

2D dD/dt = 67,600 sin(θ) dθ/dt

2(260) dD/dt = 67,600 * √15/4 * 7/130

dD/dt = 67,600/520 * √15/4 * 7/130

dD/dt = 7√15/4

dD/dt = 6.77772

Distance between the friends is changing at a rate of 6.78 m/sec