SOLUTION: a mover drives a moving truck. he leaves the familys house and travels at a steady rate of 40mph. the family leaves three fourths of an hour later following the same route. they tr
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Question 675711: a mover drives a moving truck. he leaves the familys house and travels at a steady rate of 40mph. the family leaves three fourths of an hour later following the same route. they travel at a steady rate of 60MPH. how long after the moving truck leaves will the car catch up?
i can do the equition i just dont know how to set it up. Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
In 3/4 hr mover travels (d=rt)40*(3/4)=30 mi
When they both have traveled the same distance, the car will have caught up
Let t=time that elapses 'till they have traveled the same distance
distance Mover travels=30+40t
distance car travels =60t
30+40t=60t
20t=30
t=1.5 hrs----time it takes the car to catch up after the car leaves. May need to add 3/4 hr to this if we are calculating from when the truch leaves
CK
Mover travels 30+40.(1.5)=30+60=90 mi
Car travels 60*(1.5)=90 mi
Hope this helps---ptaylor
