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Displacement and average velocity

  1. Dec 17, 2006 #1
    1. The problem statement, all variables and given/known data
    A hiker travels south a long a straight path for 1.5 h with an average velocity of .75 km/h. then travels south for 2.5 h with an average velocity of .90 km/h. What is the hiker's displacement for the total trip?


    2. Relevant equations



    3. The attempt at a solution
    For the answer I have 3.4 km to the south.
    but hm, I think I copied the question wrong. I have a feeling that I was supposed to have another coordinate sign besides south... so I'm not sure.
     
  2. jcsd
  3. Dec 17, 2006 #2

    cristo

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    Well, your answer is correct (3.375 km i get before rounding) if the question is as you've written it!!
     
  4. Dec 17, 2006 #3
    oh ok. cool.
    but oh gosh, let me try it. but well I'm not really sure what to do.
     
  5. Dec 17, 2006 #4

    cristo

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    Ahh, so you mean that's the answer the book gives? Ok, well what equation do you know relating velocity, distance and time?
     
  6. Dec 17, 2006 #5
    well its a review sheet, the teacher gave it to me.

    distance/t
    or do I use: av. velocity= total distance/time
     
  7. Dec 17, 2006 #6

    cristo

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    Yes, use av.velocity=distance/time, but remember to calculate the two parts separately, and then add them up at the end to arrive at a total distance.
     
  8. Dec 17, 2006 #7
    awesome.
    another question, for instance 1.5 h, ok do I convert that into minutes? and if so do I take the .5 as half or as 50, as in fifty minutes. 1 hour and fifty minutes or 1 and a half or 90 minutes or 110 minutes?
    wow, I ask a lot of questions.
     
  9. Dec 17, 2006 #8

    cristo

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    Well, you would take 1.5h as one and a half hours, or 1 hr 30 mins. However, in this instance, you don't need to convert.

    The units of velocity are given as km/hr, and the times in hr, so, if we look at the equation: s=vt, and consider putting the units in, we get s (km) = v(km/hr)*t(hr). Notice that the hr's will cancel, and give a result in km, which is what is required.

    For calculations like this, as long as you keep to the same units throughout, the answer will be correct. (A problem would occur if, for example, the velocity was in km/hr and the time measured in seconds. In this case we would have to convert units)

    Hope this is clear!
     
  10. Dec 17, 2006 #9
    yeah thats what I noticed.
    but well ok for km/h I get 1.65 and then for hours 4.
    so this is my equation: 1.65 km/h /4h and then I get .4125...
    eek. I know I'm doing something wrong.
     
  11. Dec 17, 2006 #10

    cristo

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    Re-read post #6. You need to do one calculation to work out the distance for the first part: travelling along a straight path for 1.5 h with an average velocity of .75 km/h, and another calculation for the second part: travelling for 2.5 h with an average velocity of .90 km/h.

    When you have these two distances, add them up and you will get the required answer.
     
  12. Dec 17, 2006 #11
    ooops, sorry. got you. thanks!
     
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