SOLUTION: two cars start 100 miles apart. the faster goes 90 mph. the slower goes 60 mph. how far from the starting location of the faster car will they meet? how much time will have passed?
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Question 246379: two cars start 100 miles apart. the faster goes 90 mph. the slower goes 60 mph. how far from the starting location of the faster car will they meet? how much time will have passed?
i tried d= rt
distance r(t)
100= (90+60)t
100/150=150/150(t)
.667=t
60 miles =d?? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! two cars start 100 miles apart. the faster goes 90 mph. the slower goes 60 mph. how far from the starting location of the faster car will they meet? how much time will have passed?
i tried d= rt
distance r(t)
100= (90+60)t
100/150=150/150(t)
.667=t
60 miles =d??
OK. YOU HAVE THE CORRECT FORMULA!!!! AND I GOT THE SAME ANSWER THAT YOU GOT!!! GOOD WORK!!! HERE'S MY SOLUTION. I LIKE TO LAY MY PROBLEMS OUT IN A STEP BY STEP FASHION SO I'LL KNOW WHERE I AM AT ANY GIVEN TIME IN THE SOLUTION.
Distance(d) equals rate (r) times time(t) or d=rt; r=d/t and t=d/r
Distance travelled by faster car=rt=90t
Distance travelled by slower car=rt=60t
Now when the above two distances add up to 100 miles the two cars will have met, so:
60t+90t=100
150t=100 or
t=2/3 hour
So distance travelled by faster car =90*(2/3)=60 mi
Distance travelled by slower car =60*(2/3)=40 mi
