Question 938719: Two angry children 470 feet apart are headed straight towards each other—one at 15 feet per second and the other at 9 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?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Two angry children 470 feet apart are headed straight towards each other—one at 15 feet per second and the other at 9 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?
------
Closing speed for the children is 15+8 = 23 ft/sec
----
Time till they meet is 470/23 = 20.43 seconds
------
Ans:: In that time the mother has gone 20*20.43 = 408.7 feet
----------------
Cheers,
Stan H.
----------------
|
|
|
