SOLUTION: paul left springfield traveling north at an average speed of 64 miles per hour. tim left springfield traveling south at an avergae spped of 58 miles per hour. how long would it tak

Algebra ->  Equations -> SOLUTION: paul left springfield traveling north at an average speed of 64 miles per hour. tim left springfield traveling south at an avergae spped of 58 miles per hour. how long would it tak      Log On


   

Question 268982: paul left springfield traveling north at an average speed of 64 miles per hour. tim left springfield traveling south at an avergae spped of 58 miles per hour. how long would it take after paul left for them to be 982 miles apart? how many hours would it be after paul left?
Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
This is an RTD problem. Here is a table based on the given information
direction . . . . . . . . .rate . . . . . . . . .. time . . . . . . . . .. distance
Paul (N) . . . . . . . . . . 64. . . . . . . . . . . t. . . . . . . . . . . . . 64t
Tim (S) . . . . . . . . . . .58. . . . . . . . . . .t - h. . . . . . . . . . .58t - 58h
distances . . . . . . . . . . . . . . . . . . . . .2t. . . . . . . . . . . . 982
We don't know when Tim left. We want them 982 miles apart but the distance are increasing, so we add to get
64t + 58t - 58h = 982
or
122t - 58h = 982
solving for t, we get
t = (982+58h)/122
If h = 2, then t = 9
It would 9 hours after PAul left to be 982 miles apart.
It would be 7 hours after Tim left.