Help me please!

Question

A pair of kids and a pair of adults decided to compete in a three-legged race. The kids got to start 15 yards ahead of the adults, since they had shorter legs. When they were told to start, the kids hobbled forward at a rate of 2 yards per second, and the adults hobbled after them at a rate of 3 yards per second. Soon they were side-by-side. How long did that take?

Answer 1

Kids travel 15 + 2s, where s represent the time in seconds. (well they do not travel 15 yards - that is their head start)

Audlts travel 3s, where s represents the time in seconds

15 + 2s = 3s

Casn you solve for s and answer?

Answer 2

Here's another approach:

D = rt, so t = D/r

When the adults catch up to the kids, both have run for the same length of time but the kids distance was 15 yds less.

t_k = (D - 15)/2; t_a = D/3, where D is the distance both have travelled when they are side-by-side.

t_k = t_a

(D - 15)/2 = D/3. Cross-multiply: 2D = 3(D - 15)

2D = 3D - 45; D = 45 yds. How long did it take?

t_k = (45 - 15)/2 = 15 sec.

t_a = 45/3 = 15 sec.