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Motion in a straight line

  1. Sep 15, 2009 #1
    1. The problem statement, all variables and given/known data

    You allow yourself 35 minutes to drive 25 miles to the airport, but are caught in heavy traffic and average 22 miles/hour for the first 15 minutes. What must your average speed be on the rest of the trip if you are to get there on time? Express your answer in miles/hour.

    2. Relevant equations

    D=vt

    3. The attempt at a solution

    This was confusing, but this is how i started. I did a little dimensional analysis and did (25mil/35min) x (60min/1hr)= 42.857 mil/hr. From what I understand, this is what you need to do to get to the airport on time. I'm going 22miles/hour for the first 15 mins, meaning, i only have (35-15) = 20 mins left to get to the airport on time. Then I did 42.857-the miles i orginally traveled, which in this case was 22, leaving me with 20.857 miles to go.

    So I used D=vt.
    20.857=(v)(20min). V=1.04mil/hour

    This would be the speed I would need to be traveling for the next 20 mins in order to get to the airport on time.

    Is this correct?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 15, 2009 #2

    rl.bhat

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    In the first 15 minutes the average velocity is 22 miles/h. Let v be the average velocity in the remaining time.
    So the net average velocity = (22+v)/2
     
  4. Sep 16, 2009 #3
    Hmm, If i'm solving for V, but how do I get that "net average velocity"?
     
  5. Sep 16, 2009 #4

    rl.bhat

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    You have already calculated it. That is total distance /total time.
     
  6. Sep 16, 2009 #5
    So if im understanding this right, total distance/total time was 25mil/35 min?, which was 42.857mi/hr? and im solving for V?

    if so, I did this: 42.857= (22+v)/2
    v=63.7mi/hr, which would be the speed I would need to go in order to make it to the airport on time.

    Is this it?
     
  7. Sep 16, 2009 #6

    rl.bhat

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    I think so.
     
  8. Sep 16, 2009 #7
    ouch, apparently it was wrong lol.

    I dunno, I think I'm supposed to use the "22mil/hour for 15 min" data somehow, perhaps a bit of conversion

    any other ideas?
     
  9. Sep 16, 2009 #8

    rl.bhat

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    Try this one.
    In 15 minutes he moves 5.5 miles
    Find the remaining distance and remaining time. Then find the average velocity for the second part.
     
    Last edited: Sep 16, 2009
  10. Sep 16, 2009 #9
    How did you get 5.5 kilometers? Did you use a conversion factor between kilomters and miles from 22 miles? And why exactly am I using kilometers?
     
  11. Sep 16, 2009 #10

    rl.bhat

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    Sorry.It is miles.
    22 mile*15 min/60 min
     
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