Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Uniformly accelerated motion- a weird kangaroo prob

  1. Sep 28, 2004 #1
    Uniformly accelerated motion-- a weird kangaroo prob

    Ok, I have this simple Question:

    A kangaroo can jump straight up 2.5 m -- what is its takeoff speed?
    How can I solve that given just one variable. Well, I know that Vi=0, right?

    Also, if you anyone can help me try to solve this one?:

    A car in an auto accident left skid marks 290 m long. Assuming an average acceleration of -3.9 m/s/s (that is, -0.4 g), calculate the Jag's speed when the brakes locked.

    I know acceleration and scalar distance, but I don't know anything about the initial speed and such. So how can I choose a formula?

    Please Help.

    Thanks
     
  2. jcsd
  3. Sep 28, 2004 #2

    Pyrrhus

    User Avatar
    Homework Helper

    For first problem, review Free Fall.

    For the second problem, what happens when you apply the brakes?
     
  4. Sep 28, 2004 #3
    When the brakes applied, the car slows, right? And skids 290 m. I'm not sure where to go from there :confused:
     
  5. Sep 28, 2004 #4
    This is how i believe the question is solved:
    Vi=?
    Vf=0m/s
    a=-3.9m/s^2
    t=X
    d=290 m

    use thsi formula
    Vf^2 = Vi^2 + 2ad
     
  6. Sep 28, 2004 #5
    I'm an Aussie, so I should be able to help with the Kangaroo question! :biggrin:

    First off, you have more than one variable.

    (Assuming that the kangaroo is on the Earth) The acceleration can be given by g. (ie use -9.8 m/s/s)

    The displacement is 2.5m, and Vf = 0 (at the top of the jump, the kangaroo's velocity is zero).

    Vi is the unknown, that you want to calculate.

    I believe the constant acceleration formula you might need is:

    [tex] v_f^2 = v_i^2 + 2as[/tex]

    where Vf and Vi are the final and initial velocities respectively.
    a is the acceleration
    s is the displacement


    also don't forget the direction of things. (if you use +2.5 for the displacement, you MUST use -9.8 for the acceleration)
     
  7. Sep 29, 2004 #6
    thanks

    Great, thanks Tyco and punjabi. I got the problems, and I actually understand them. :biggrin:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Uniformly accelerated motion- a weird kangaroo prob
Loading...