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
