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Homework Help: Objects moving in 2D

  1. Dec 5, 2006 #1
    1. The problem statement, all variables and given/known data

    the coordinates of an object moving in the 2D space are given by x=2t and y=4t^2. what are the dimensions of the numerical coefficients 2 & 4, respectively? What are the objects velocity, speed and acceleration and the magnitude of the acceleration at t=0. what is the angle between the velocity and the acceleration at t=0

    2. Relevant equations

    have been through my text book god knows how many times and cannot decide which to use

    3. The attempt at a solution

    i have honestly tried for hours and havent managed to get anywhere

    however, for the dimensions i substituted 2 and 4 in respectively, but i doubt this is correct
    Last edited: Dec 5, 2006
  2. jcsd
  3. Dec 5, 2006 #2
    I think you meant y = 4t^2 in the second equation.

    For the fist equation, what are the dimensions of x and t. Having found that, what should be the dimension of 2, for the equation to be valid?
  4. Dec 5, 2006 #3
    edited, thanks

    i got :

    x=2 = 4
    x=4 = 8

    y=2 = 64
    y=4 = 256

    however, like i said, i doubt this is correct....or is it?
  5. Dec 5, 2006 #4
    with those equations, should you put them together to get a y=mx+c styled equation?

    i get y = x^2 and then go from there maybe?
  6. Dec 5, 2006 #5


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    Homework Helper

    You can write the following: [tex]\vec{r}(t) = x(t)\vec{i}+y(t)\vec{j}[/tex]. Plug the given functions in, and use the following relations: [tex]\frac{d\vec{r}(t)}{dt}=\vec{v}(t)[/tex], [tex]\frac{d\vec{v}(t)}{dt}=\vec{a}(t)[/tex].
  7. Dec 5, 2006 #6
    Nope, it's not.

    The DIMENSION of x is Length(L) and the dimension of t is Time(T). Now, what is the coefficient's dimension to make a physically meaningful equation?

    Whichever textbook you may be using, the first or second chapter probably covers the topic of units and dimensions. Give it a read.
  8. Dec 5, 2006 #7
    its adding vectors?

    i think for x(t)

    you put 2 into x=2t, so you get 4?

    then for y( t)

    you put 4 into y=4t^2, so you get 74 (4x16)

    so r( t) = 4i + 74j ?
    Last edited: Dec 5, 2006
  9. Dec 5, 2006 #8


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    Science Advisor

    Is WHAT "adding vectors"? Which question are you answering here?
    t is the TIME, measured, say, in seconds. Why should the time be 2 seconds for x and 4 seconds for y? Those equations give the (x,y) coordinates of the point at time t seconds.

    The first question, which I think you are trying to answer, asks "the coordinates of an object moving in the 2D space are given by x=2t and y=4t^2. what are the dimensions of the numerical coefficients 2 & 4, respectively?
    What kind of UNITS should they have, not numerical values. In x= 2t, t will have TIME units, x will have DISTANCE units. Think of it as multiplying fractions: (Distance/Time)(Time)= Distance. That "2" factor must have units of Distance/Time (so it's really a "speed" factor), for example meters/sec if t is measured in seconds and x in meters, or miles/hour if x is measured in miles and t in hours.

    Now look at y= 4t^2. Again, t is a TIME, y is a DISTANCE. t^2 will be Distance^2. To get Distance= ( )(Time^2), that ( ) must look like Distance/Time^2 so that the "Time^2" terms cancel and leave "Distance". The "4" must have units of Distance/Time^2 (so it's really an 'acceleration' factor), for example meters per second per second or miles per hour per hour.
  10. Dec 5, 2006 #9
    thanks, that was a very useful post

    i made sense of that!

    for the velocity, how would you get a direction on the force?

    also can you reach an answer or should the answer be displayed in equation format?

  11. Dec 5, 2006 #10


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    Staff Emeritus
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    Gold Member

    If you go back to post number 5 by radou he explains exactly how to complete the question. The speed is the magnitude of the velocity and is given by using pythagoras' theorem.

    [tex] |\mathbf{v}|= \sqrt{x(t)^2+y(t)^2} [/tex]

    Of course its made rather simple by the fact that its all at time t=0.
    Last edited: Dec 5, 2006
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