Two cars travel in the same direction along a straight highway, one at a constant speed of 57 mi/h and the other at 76 mi/h. Assuming they start at the same point, how much sooner does the faster car arrive at a destination 12 mi away? Answer in units of min. How far must the faster car travel before it has a 20 min lead on the slower car? Answer in units of mi.
Accepted Solution
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Answer:Car 2 arrives 3.16 minutes sooner to car 1 at 12 miles.The faster car must travel 75.76 miles before it has a 20 min lead on the slower car.Step-by-step explanation:Speed of car 1 = 57 mphSpeed of car 2 = 76 mphWe need to find how much sooner does the faster car arrive at a destination 12 mi away.Time taken for car 1 [tex]t_1=\frac{12}{57}=0.21hour=12.63minutes[/tex]Time taken for car 2 [tex]t_2=\frac{12}{76}=0.158hour=9.47minutes[/tex]Difference between arrival = 12.63 - 9.47 = 3.16 minutesSo the car 2 arrives 3.16 minutes sooner to car 1 at 12 miles.Now we need to find the distance at which faster car has a 20 minute lead Difference between arrival = 20 minutesLet the distance be STime taken for car 1 [tex]t_1=\frac{S\times 60}{57}=1.053S[/tex]Time taken for car 2 [tex]t_2=\frac{S\times 60}{76}=0.789S[/tex] We have 1.053 S - 0.789 S = 20 S = 75.76 milesSo the faster car must travel 75.76 miles before it has a 20 min lead on the slower car.