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Homework Help: Distance, displacement

  1. Mar 19, 2012 #1
    1. The problem statement, all variables and given/known data

    A particle's position is described by x=2t^3-4t-5
    Find its a) displacement b) distance
    from t=0 to t=5

    2. Relevant equations

    3. The attempt at a solution

    Displacement, plug in x(5)-x(0) = 230m
    Distance, shouldn't it be the same thing plug in and get 230m? But book says 234m. No idea how this came about.
  2. jcsd
  3. Mar 19, 2012 #2
    Hey ,

    Do you know the difference between distance and displacement (Am sure you do! :-))

    If a particle goes from x = 2m to x=4m and then comes back to x=3m

    What is the distance travelled and displacement?
    How did you calculate it in simple terms.
    Something similar is happening here.
  4. Mar 19, 2012 #3
    distance would be the actual physical length, so that would be 3m. Displacement is just the direct distance from start to finish, so that would be 1m.
  5. Mar 19, 2012 #4
    Yes.So the basic logic you used is that you added the path length.
    When the body reversed its direction at 4m ( in physics what is this point signify in terms of variables?) you didn't add the path from 4m to 3m as -1m.You added it as 1m.

    Do the same on this question and you will get the answer:-)
  6. Mar 19, 2012 #5
    alright man but ur concept was a simple question, this one has complicated equations and i don't know how to apply it : ======================== (
  7. Mar 19, 2012 #6
    See displacement is integration of (vdt).
    Distance is integration of (|v|*dt)
    And |v| is speed.

    For solving the mod || you need to find intervals where v is positive and where v is negative.
  8. Mar 19, 2012 #7
    so take absolute value of x=2t^3-4t-5.

    then it would be -2t^3+45+5
    so i take the integral of this and that's it?
  9. Mar 19, 2012 #8
    For what value of t is the velocity positive?
    When is it negative ?

    take the negative of v where velocity is negative.

    Take positive when velocity is positive.
    Then integrate.
    (Why do we do so?)
  10. Mar 19, 2012 #9
    ok, so when x=0 is the difference. So x is negative the function is: -2t^3+4t+5, and when x is positive the function is: 2t^3-4t-5

    add these two up and integrate?
    but if i add it up i get 0?
  11. Mar 19, 2012 #10
    No.We are interested in mod of velocity not x.take function as it is when v is positive(not neccesarily x) and multiply by minus when v is negative (not x)
  12. Mar 19, 2012 #11
    No.We are interested in mod of velocity not x.take function as it is when v is positive(not neccesarily x) and multiply by minus when v is negative (not x).

    And you wont get 0.
  13. Mar 19, 2012 #12
    hey can you just show me the process? I'm getting very confused and I'm not sure what's going on. I think I can understand better if you show me the math. velocity would be x'=6t^2-4

    Thanks man.
  14. Mar 19, 2012 #13
    You calculated the velocity correctly. :-)

    Property of mod function is that
    when f(x) is positive

    |f(x)| =f(x)

    when f(x) is negative

    |f(x)| =-f(x).

    For eg if |2| is 2.|-2| is -(-2) i.e 2 again.


    Velocity is 6t^2-4.
    which is positive when t belongs from sqroot (2/3) to 5 and negative between 0 to sqroot(2/3.)
    So take displacement of [-(2t^3-4t-5)] for t = 0 to t=sqroot(2/3).
    (or you can integrate

    And take displacement of [(2t^3-4t-5)] for t=sqroot(2/3) to t=5.
    (or you can integrate
  15. Mar 19, 2012 #14
    how did you get sqrt(2/3)??
    I guess that's from x=2t^3-4t-5??
    so 0=2t^3-4t-5
    But how do I solve this cubic equation???

    Thanks a lot man.
  16. Mar 19, 2012 #15
    From 6t^2 -4 >0
    We are interested in |v| not mod x.

    See in the example i gave you .

    From 2 to 4 m velocity was positive so you did 4-2 =2
    (final position -initial position)

    Then particle turned behind and velocity became negative.in moving from 4 to 3 you therefore did -(3-4)=1

    And added these.

    All this while x was positive.

    Our sense of mod depends on velocity and not position :-)

    My pleasure :-)

    I hope am making stuffs clear enough :-)
  17. Mar 19, 2012 #16
    thanks a lot man europeans are superior
  18. Mar 19, 2012 #17
    Lol. I am an Indian :-)
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