2440.71 – Uther and Eethyl

Uther can run at a constant speed of kk yards per second. Eethyl is faster: she can run xx times as fast as Uther, xx>1. Suppose Eethyl gives Uther a head start of yy yards, and then they both start running at the same time in the same direction. How many yards will Eethyl run before she catches up with Uther? Your answer will be some kind of a formula involving the variables in this problem.

Solution

Filling in the chart: Let dd = how far Eethyl runs, and let d−yd - y equal Uther's distance. The times for the two are equal.

So t=t:d−yk=dxk→dk=xk(d−y)→dk=xkd−xky→xkd−dk=kxy→=d(xk−k)−kxy→d=kxyxk−k=kxyk(x−1)=xyx−1t = t: \frac{d-y}{k} = \frac{d}{xk} \rightarrow dk = xk(d-y) \rightarrow dk = xkd - xky \rightarrow xkd - dk = kxy \\ \rightarrow = d(xk - k) - kxy \rightarrow d = \frac{kxy}{xk-k} = \frac{kxy}{k(x-1)} = \frac{xy}{x-1}(seconds).

2440_71_solution_bdcc3bff52.png