Steve and Elsie are camping in the desert, but have decided to part ways. Steve heads north, at 8 AM, and walks steadily at 2 mi

Question

3 months ago

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Steve and Elsie are camping in the desert, but have decided to part ways. Steve heads north, at 8 AM, and walks steadily at 2 mi

les per hour. Elsie sleeps in, and starts walking west at 2.5 miles per hour starting at 10 AM. When will the distance between them be 25 miles?

Physics

1 answer:

Keith_Richards [3.2K] · Answer 1 · 2026-03-07T10:56:50+03:00

Answer:

2.57 hours

Explanation:

Let t (in hours) represent the time it takes for Elsie to walk until they are separated by 25 miles. Since Steve starts 2 hours earlier, his time will be t + 2.

The distance covered by Steve heading north is $s_s = 2(t + 2)$

The distance Elsie travels heading west is $s_e = 2.5t$

The separation between Steve and Elsie is

$\sqrt{s_s^2 + s_e^2} = \sqrt{(2(t+2))^2 + (2.5t)^2} = 25$

We can solve for t by squaring both sides.

$(2(t+2))^2 + (2.5t)^2 = 25^2 = 625$

$4(t+2)^2 + 6.25t^2 = 625$

$4(t^2 + 4t + 4) + 6.25t^2 = 625$

$10.25t^2 + 16t - 609 = 0$

$t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$t= \frac{-16\pm \sqrt{(16)^2 - 4*(10.25)*(-109)}}{2*(10.25)}$

$t= \frac{-16\pm68.74}{20.5}$

Therefore, t = 2.57 or t = -4.13.

Since t cannot be negative, we select t = 2.57 hours.