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Homework Help: Instantaneous velocity

  1. Aug 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Determine the instantaneous velocity at time t = 4, 0 s


    2. Relevant equations

    3. The attempt at a solution

    I believe it has something to do with the derivative of the point at t = 4.0s but I don't know how to solve this. I need some guidance.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Aug 21, 2013 #2


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    What is the graphical representation of a derivative?
  4. Aug 21, 2013 #3


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    Hint : Think about the slope of a straight line.
  5. Aug 22, 2013 #4
    I have absolutely no idea, I'm more lost than I would be in the rain forest (whot did I just make a bad joke? Yes I did)..

    Seriously I have a headache, been stuck on several other problems as well and the ones I have solved I am starting to get really paranoid about and questioning wether or not it's right. Overthinking every aspect of those problems even though they should be easy. Went to sleep 7hrs ago, woke up 1 hr ago thinking I had figured out what's wrong with my previous calculations and couldn't go back to sleep so I started working on it again only to find I still don't know. fml
  6. Aug 22, 2013 #5


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    The equation for a straight line is y=mx + c where m is the slope and c is a constant.

    On your graph y is distance in meters, x is time in seconds so the slope is the velocity.

    The slope of the line between two points (x1, y1) and (x2, y2) can be calculated from (y2-y1)/(x2-x1) or in English.. the change in y divided by the change in x... or the change in distance divided by the change in time.
  7. Aug 22, 2013 #6
    Okay, so if I should take the derivative of t = 4.0 it should be x=4.0 y=12.5

    so 12.5 / 4 = 3.125 so the instantaneous velocity is ≈3.1m/s?
  8. Aug 22, 2013 #7


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    that would be correct if the slope was constant throughout. But in this case it is not. You need to calculate the gradient of the slope at t = 4.
  9. Aug 22, 2013 #8
    Like this?: if I calculate t=3 I get 10/3 = 3.33m/s. if I take t=5 i get 15/5 = 3m/s

    (15 - 10) / (5 - 3) = 5/2 = 2.5. is it 2.5m/s?

    I am sorry but I'm really not too sharp at the derivative.
    Last edited: Aug 22, 2013
  10. Aug 22, 2013 #9


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    yes, that is the right answer :)
    The way I like to think of it is like this: between t=3 and t=5, the person (or whatever it is) moves 5 meters, so he moves 5 meters in 2 seconds. So during this time, his average velocity is 2.5m/s and since his velocity is the same during this time, this is also equal to his instantaneous velocity.
  11. Aug 22, 2013 #10
    Thank you for your help sir :) Really do appreciate it :)

    That's a good thought, how would you solve a more complicated but similar problem? Like if it's a curve or 2 curves that stretches over t=3 and t=5 but instead the point that we just now calculated would be on t=3.3 in a curve instead.

    PS: not necessary with an answer, just something I thought about.
  12. Aug 22, 2013 #11


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    If it's a curve, you would need to find the slope of the tangent line. The slope of the tangent line is (calculus terminology) the limit of the difference quotient. In calculus you will learn about this.
  13. Aug 22, 2013 #12
    its easier than you think...just draw the tangent to the curve and calculate the slope. Dont forget to use the units on the axes.
  14. Aug 22, 2013 #13
    Okay, thank you guys :)
  15. Aug 23, 2013 #14
    This answer I got apperently was wrong.

    (15 - 10) / (5 - 3) = 5/2 = 2.5. is it 2.5m/s is the wrong answer for this problem.

    My teacher said to find out the Instantaneous velocity in t=4.0s I should do it like this. In the time range 2.5 s to 7.0 s, the speed is constant. Instantaneous velocity at t = 4.0 s is therefore equal to this constant speed.

    We determine this by the slope. v = m / s = 1.7 m / s.

  16. Aug 23, 2013 #15


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    Bruce in post #9 gave you the correct method, but (not surpisingly) could not read accurate values of displacement from your sketch graph.
    While you can take the slope from the complete straight section of the graph - from time 2.5 -> 7.0, you can always take the slope from any part of that straight section.
    As I said, the nature of the graph you posted makes it rather difficult to estimate actual values of displacement at various times. The real graph you were working from was presumably much easier to read accurately.
  17. Aug 23, 2013 #16


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    ah that's unfortunate. your method is correct. The speed is constant from 2.5 s to 7.0 s So this also means the speed was constant from 3 s to 5 s. The reason your answer was different is only because the graph is quite hard to judge.

    Although, on the other hand, it is good practice to use a longer time interval over which the speed is constant. (Because this makes it easier to judge what the slope is). So the time interval from 2.5 s to 7.0 s is the 'ideal', since the part of the graph you are looking at is as large as possible, while still being over a section of constant speed.
  18. Aug 23, 2013 #17
    That's the very same graph I got :P

    But thank you for your comment :)
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