Stella, in spite of her age, still jogs regularly and can keep up a speed of k meters per second. Her young friend Jasper can jog at a speed x times as fast, with x > 1 (definitely). Suppose Jasper gives Stella a head start of y meters, and then they both start jogging at the same time in the same direction. How many meters will Jasper jog before he catches up with Stella? Your answer will be some sort of formula involving the variables in this problem.
| d = | r | t | |
| J | d | xk | t |
| S | d – y | k | t |
Filling in the chart: Let d = how far Jasper runs, and
d – y = Stella’s distance. The times for the two are equal.
d – y = Stella’s distance. The times for the two are equal.
So t = t, that is, (d – y)/k = d/xk
→ dk = xk(d – y)
→ dk = xkd – xky
→ xkd – dk = kxy
→ xdk – dk = kxy
→ d(xk – k) = kxy
→ d = kxy/(xk – k)
→ d = kxy/(k(x – 1))
→ d = xy/(x – 1).
→ dk = xkd – xky
→ xkd – dk = kxy
→ xdk – dk = kxy
→ d(xk – k) = kxy
→ d = kxy/(xk – k)
→ d = kxy/(k(x – 1))
→ d = xy/(x – 1).