The anticipated time when he will appear again is 5:17 pm
The given parameters;
your speed, [tex]V_a[/tex] = 3 m/s
the runner's speed, [tex]V_b[/tex] = 14 m/s
the diameter of the lake, d = 2 miles = 3218.69 meters
The circumference of the lake is calculated as;
[tex]C = \pi d\\\\C = 3.142 \times 3218.69 = 10,113.12 \ m[/tex]
Apply concept of relative velocity to determine the time, in which he will appear again.
By the time he appears again;
the distance you moved + distance he moved = circumference of the circle
[tex]V_at + V_bt = 10, 113.12\\\\(V_a + V_b)t = 10,113.12\\\\(3 + 14) t = 10,113.12\\\\17t = 10,113.12\\\\t = \frac{10,113.12}{17} \\\\t = 594.89 \ s = 9.92 \ \min \ \approx 10 \ \min[/tex]
[tex]t\ \approx \ \ 5:17 \ pm[/tex]
Thus, the anticipated time when he will appear again is 5:17 pm
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