Solving Kinetics Questions: Two Cars Traveling on a Highway

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Two cars are traveling on a highway at speeds of 55 mi/h and 65 mi/h, respectively. For part (a), the faster car arrives at a destination 10 miles away 10.91 minutes sooner than the slower car. To solve part (b), the discussion emphasizes using the relationship between speed, distance, and time to determine how far the faster car must travel to achieve a 15-minute lead. The solution involves adjusting the distance based on the time difference established in part (a). The conversation concludes with the user gaining clarity on the problem-solving approach.
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Homework Statement


Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 65 mi/h.

(a) Assuming that they start at the same point, how much sooner does the faster car arrive at a destination 10 mi away?
______ min
(b) How far must the faster car travel before it has a 15-min lead on the slower car?
______ mi


Homework Equations


I'm not sure how to set up an equation.


The Attempt at a Solution


I divided 10 mi by each speed, then multiplied each result by 60 and compared minutes.
 
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For part a you seem to have started off correctly. All you need to do is find the difference in arrival times.

Any ideas for part b?
 
Welcome to PF, aquapod17.

You did part (a) correctly. The relevant equation is the one that relates speed, distance, and time to each other. (You must have been told this equation, or it's in your textbook ... otherwise you would not be asked to do a problem like this.)

(b) Instead of 10 miles, what would the distance be so that your answer is 15 minutes? You already know, from part (a), if that distance is greater than or less than 10 miles.
 
Thanks I got it!
 
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