SOLUTION: Two cars left daisy's Diner at the same time and traveled in opposite directions. One car traveled 78 minutes. the other car traveled for 144 min at a rate of 5kmh faster than the
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Question 67695This question is from textbook
: Two cars left daisy's Diner at the same time and traveled in opposite directions. One car traveled 78 minutes. the other car traveled for 144 min at a rate of 5kmh faster than the first car. If the faster car went twice as far as the slower car, how far did each car travel? (express the times in hours) This question is from textbook
distance(d)=rate(r) times time(t) or d=rt
Distance first car travelled=78x=1.3x (expressed in hours)
Distance second car travelled =144(x+5)=2.4(x+5) (expressed in hours)
Now we are told that the faster car went twice as far as the slower car. So our equation to solve is:
2.4(x+5)=2(1.3x) clear parens
2.4x+12=2.6x subtract 2.4x from both sides
0.2x=12
x=60 kmh--speed of first car
1.3x=1.3(60)=78km-------------------------distance travelled by 1st car
x+5=60+5=65 kmh speed of other car
2.4(x+5)=2.4(65)=156 km -------------------distance travelled by other car
ck
156=2(78)
156=156
