Question

Two angry children 450 feet apart are headed straight towards each other—one at 15 feet per second and the other at 6 feet per second. Their mother responds to the crisis by sprinting back and forth between the children, hoping to calm them before they reach each other. Unfortunately, when she reaches each child, she is immediately rejected and rushes back to the other child. If the mother miraculously maintains a constant speed of 20 feet per second the entire time—while racing back and forth between the children—how far will she have run when all three of them collide? She will have run __________ feet. (Round to the nearest foot.)

Answer 1

If the amount of time is x seconds, then 15*x is the amount ran by one child and 6*x is the amount ran for the other child due to that they add 15 or 6 feet for each second. In addition, since they are 450 feet apart, 15x+6x=450=21x. Dividing both sides by 21, we get x=around 21.4 seconds. Since the mother is always between them, when those two meet is when the three meet. Therefore, she runs 20* around 21.4= around 428.57 feet