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anonymous

  • one year ago

My friend sets out walking at 3 miles per hr. I set out behind her 5 minutes later at 4 miles per hour. For how many minutes will my friend have been walking when I catch up with her?

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  1. anonymous
    • one year ago
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    I think that you need to do 3*1/12=1/4 mile which is the distance between them after 5 minutes.

  2. anonymous
    • one year ago
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    then time it takes for them to catch up by 1/4 mile is 12

  3. anonymous
    • one year ago
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    is it not?

  4. anonymous
    • one year ago
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    because every hour there is a disparity of 1 mile between them.

  5. anonymous
    • one year ago
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    so for them to cover 1/4 mile they would need 1/4 hour .

  6. anonymous
    • one year ago
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    60*1/4=15

  7. anonymous
    • one year ago
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    yeah sorry about that it's I think 15min your thoughts?

  8. kropot72
    • one year ago
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    Let d be the distance walked before catch up. Time in minutes taken by friend at catch up = d/3 * 60 Time in minutes taken by me at catch up = d/4 + 5 Therefore equating the times we get \[\large \frac{d}{3}\times60=\frac{d}{4}\times60+5\] Now if you solve to find the value of d, you can plug the d value into d/3 * 60 to find the time to catch up in minutes.

  9. anonymous
    • one year ago
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    Since the slower guy with 3 mile/per hour walked for 5 minutes, it's 3*1/12=1/4

  10. kropot72
    • one year ago
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    d turns out to be 1 mile. Therefore the required time is 20 minutes.

  11. kropot72
    • one year ago
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    @danyboy9169 Do you understand?

  12. anonymous
    • one year ago
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    I don't think so

  13. anonymous
    • one year ago
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    By the time 4 mile guy starts off there is a 1/4 mile distance

  14. kropot72
    • one year ago
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    What part don't you understand?

  15. anonymous
    • one year ago
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    because 5/12=1/12 1/12*3miles/hour=1/4 mile between them

  16. UsukiDoll
    • one year ago
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    we're using distance formula aren't we ?

  17. anonymous
    • one year ago
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    It's physics

  18. anonymous
    • one year ago
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    1/4mile to be covered with 1mile/h is of course it takes 1/4h

  19. anonymous
    • one year ago
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    60*1/4=15min

  20. anonymous
    • one year ago
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    Or if you taken the extra 5 min from the friend it will be 20min

  21. anonymous
    • one year ago
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    is it not?

  22. kropot72
    • one year ago
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    "Time in minutes taken by me at catch up = d/4 * 60 + 5" Correction to my first post. The rest is correct.

  23. UsukiDoll
    • one year ago
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    might as well create a table Friend - 3 mph (speed) me - 4mph (speed) 5 minutes later so there's a delay

  24. anonymous
    • one year ago
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    It's classical kinematics physics question

  25. UsukiDoll
    • one year ago
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    1 hour = 60 minutes.

  26. anonymous
    • one year ago
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    You convert 5 min to hour. 5min*1hour/60min=

  27. UsukiDoll
    • one year ago
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    so we're solving for the distance

  28. anonymous
    • one year ago
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    units "min" cancel out, leaving out the 1/12h.

  29. UsukiDoll
    • one year ago
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    mhm always remember those labels XD

  30. anonymous
    • one year ago
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    Everything makes sense with units.

  31. anonymous
    • one year ago
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    3mile/hour*1/12h=3/12mile(since h cancels out) or 1/4mile

  32. anonymous
    • one year ago
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    There is 1/4mile between the friends

  33. UsukiDoll
    • one year ago
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    \[\large \frac{d}{3}\times60=\frac{d}{4}\times60+5\] \[\frac{60d}{3}=\frac{60d}{4}+5\] \[20d=15d+5\] \[20d-15d=5\] \[5d=5\] \[d=1\]

  34. anonymous
    • one year ago
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    4mile/hour-3mile/hour=difference of 1mile/hour

  35. anonymous
    • one year ago
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    for 1 mile/hour to catch up for 1/4 mile

  36. kropot72
    • one year ago
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    Let d be the distance walked before catch up. Time in minutes taken by friend at catch up = d/3 * 60 Time in minutes taken by me at catch up = d/4 * 60 + 5 Therefore equating the times we get \[\large \frac{d}{3}\times60=\frac{d}{4}\times60+5\] which simplifies to \[\large 20d=15d+5\] therefore d = 1 mile The time taken for the friend to walk 1 mile at 3 mph = 20 minutes.

  37. anonymous
    • one year ago
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    d=1/4(miles) v=1mile/h

  38. UsukiDoll
    • one year ago
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    so going back a bit if d = 1 then both sides would be 20 1 mile 20 minutes

  39. anonymous
    • one year ago
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    Use the kinematics formula t=d t=0.25mile/1mile/h=1/4h=15 minutes. Plus that 5 min for initial dilation. 15min+5min=20min

  40. anonymous
    • one year ago
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    Use physics formula it's much easier

  41. kropot72
    • one year ago
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    @Robert136 "Use the kinematics formula t=d" A new one on me!!!!!!

  42. UsukiDoll
    • one year ago
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    :P nah prefer da mathz

  43. kropot72
    • one year ago
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    How about \[\large time=\frac{distance}{speed}\]

  44. anonymous
    • one year ago
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    Yup

  45. anonymous
    • one year ago
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    You can manipulate the formula without thinking much. Only the algebra skills required

  46. UsukiDoll
    • one year ago
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    yesssssssssssssssss @kropot72

  47. anonymous
    • one year ago
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    So any missing value will be compensated by rearranging the formula

  48. anonymous
    • one year ago
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    Got it, Since I walk 1 mile per hour faster than my friend, the distance between us shrinks at a rate of 1 mile per hour. Since I am 1/4 mile behind when I start walking, it takes me 1/4 hour (= 15 minutes) to close the 1/4-mile gap. At that time, my friend, who started 5 minutes before me, will have been walking for 5+15 =20 minutes.

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