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: 2D motion problem

  1. Sep 16, 2008 #1
    1. The problem statement, all variables and given/known data
    A particle moving at a velocity of 5.2 m/s in
    the positive x direction is given an accelera-
    tion of 5.3 m/s2 in the positive y direction for
    9.9 s.
    What is the final speed of the particle?
    Answer in units of m/s

    3. The attempt at a solution
    I multiplied the acceleration by the time to get velocity.
    I found the resultant velocity between the two vectors which was 52.7. Is that the same as the final speed?
  2. jcsd
  3. Sep 16, 2008 #2
    We're looking at a scenario where object's original speed in the y-direction is 0. Then they accelerated it in the y-direction, so the final speed will represent a line going diagonally through the first quadrant of a Cartesian coordinate system.
    Acceleration is defined as change of speed over a period of time. So basically:
    a = (vf - vi)/(change in t)
    vf = at + vi = 5.3 m/s2 * 9.9s + 0 m/s = 52.47m/s <= in y-direction.

    Speed in x-direction is 5.2m/s. So its a triangle with one side being 52.47m/s and the other one 5.2m/s. Using regular Pythagorean theorem, the resultant speed (the hypotenuse, essentially) is 52.73m/s. According to significant digits, the right answer should be 53m/s.

    So what you found is indeed the right answer. Good job!
  4. Sep 16, 2008 #3
    thanks Melawrghk
  5. Sep 16, 2008 #4
    A particlemoves in the xy plane with constant
    acceleration. At time zero, the particle is at
    x = 6 m, y = 1.5 m, and has velocity ~vo =
    (3.5 m/s) ˆı + (−6 m/s) ˆ . The acceleration is
    given by ~a = (5.5 m/s2) ˆı + (6 m/s2) ˆ.

    What is the magnitude of the displacement
    from the origin (x = 0 m, y = 0 m) after
    2.5 s? Answer in units of m

    I was wondering if you have a unit vector 17.25i+9j and a starting vector of 6.18466m of the displacement vector. how could you find the magnitude of the displacement after 2.5 sec.

    I found the meters off of the resultant vector by multiplying the resultant by 2.5. Then i found the magnitude of the displacements. Where did i go wrong?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook