1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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


    User Avatar
    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


    User Avatar
    Staff Emeritus
    Science Advisor
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook