Calculating Distance for a 15 Minute Lead: Kinematics Question Explained

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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 70.0 mi/h. How far must the faster car travel before it has a 15 minute lead on the slower car?


Homework Equations



average velocity=change of position/change in time

The Attempt at a Solution



Change of position= equals average velocity multiplied by change in time

I need to know what distance the slower car covers in 15 minutes.
55mph*.25h= 13.75 miles. In other words, if the faster car was 13.75 miles ahead, it would take me 15 minutes or .25 hours to catch up.

This is what I think the answer should be, but according to the answer sheet, it's actually 64 miles. What did I miss?
 
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The question is asking how far the faster car has traveled from the beginning. So, how far will the faster car travel overall before it is 13.75 miles ahead of the slower car?
 
Ohhh! So in other words, how much distance has the faster car covered at the point where it is 13.75 miles ahead of the other car?
 
Just imagine it is just like a race.
At finishing point, faster car lead by 15 mins.

Let the time taken by faster car as x.
From this it takes slower car additional 15min to reach same distance.

From the time calculated you can find distance travelled(length of the track).