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Person $A$ starts to walk from A to B at 4 miles per hour, and on the way is overtaken by person $B$; if $B$ had started an hour later, $A$ could have walked 8 miles more before being overtaken. Find $B$'s speed.

I have no idea how to solve this.

What I do know is that when A is overtaken by B, the distances they will have traveled will be equal. Since $d=rt$, A's rate multiplied by how long he runs for until he is overtaken by B (the distance for which he walks) will be equal to B's rate times the time for which B walks. Letting x equal the time for which A walks until he is overtaken, then 4x is the distance A has covered until B catches up to him.

Applying the same to B by letting his rate equal y and the time for which he has walked until catching up to B, B's distance is yz. So 4x=yz. However, I am certain my reasoning is probably flawed and this is all wrong. But, if not, I have no clue as to how to use the other information given in the problem (the information about B starting later and A being able to walk 8 more miles until being overtaken).

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  • $\begingroup$ The distance that person A and B walk will be the same at the point where person B overtakes person A. This will also hold true in the second case (where B starts an hour later), the distances will still be the same for both person A and B here. Therefore, you set up $speed=distance$ / $time$ equations for person A and B in both scenarios, then you can eliminate the distance terms for a simplified set of equations to solve. $\endgroup$
    – FD_bfa
    Apr 10 at 3:49
  • $\begingroup$ I recommend sketching a distance-time graph and sketching on 3 lines (one for person A, one for person B, one for person B when they are 1 hour late). This will help to provide you with a visual intuition for the problem $\endgroup$
    – FD_bfa
    Apr 10 at 3:52
  • $\begingroup$ @FD_bfa the thing is that I don’t know how to get the times. For example, it says, on the way A is overtaken by B, but this doesn’t convey how long they traveled for or how much distance they covered. $\endgroup$
    – UserM1
    Apr 10 at 12:20

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