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Initial velocity for a leaping cat

  1. May 18, 2015 #1
    1. The problem statement, all variables and given/known data

    Hello, usually when solving basic velocity problems, if the object starts from rest, then its initial velocity would be 0. But why is it that, if a cat were to leap from rest, its initial velocity wouldn't be 0, but something else? My friend told me about muscle contractions being too fast or something, but I didn't really get it.

    2. Relevant equations

    Constant acceleration along a straight line equations.

    3. The attempt at a solution
  2. jcsd
  3. May 18, 2015 #2
    It really depends on the type of problems. If you're saying that a cannon is fired, you may say that the initial velocity of the cannonball is 700 m/s. If you're saying that you drop a cat off a cliff, the initial velocity is zero, and it accelerates as it falls. Does that make sense?
  4. May 18, 2015 #3
    Yes, I understand that.

    But let's say if a car accelerates from rest, its initial velocity would be 0, why is this different with the cat?
  5. May 18, 2015 #4
    It is 0, but I'm pretty sure it's a different type of problem. I think you're looking at the launch rather than the acceleration of the cat. The acceleration occurs very quickly, a very high acceleration. It accelerates very quickly to a peak velocity, and I think because the peak velocity is reached so quickly, you can ignore the infinitesimally small acceleration time.
  6. May 18, 2015 #5
    The problem is "A cat leaps upwards when startled for 0.5m in 0.3s (numbers are made up), what is its initial velocity as it leaves the ground?

    I assumed it would be 0 because the cat jumps from rest, but I checked the answer and it wasn't 0.

    Does the "as it leaves the ground" make a difference? Is that why it's not 0? So there are two initial velocities, before it leaves the ground (acceleration begins), and as it leaves the ground (but I wouldn't really call this initial velocity?)

    Is it correct to say that the problem doesn't ask for initial velocity, since that would be 0 because the cat jumps from rest, but rather the velocity at the start of the motion.

    If I were to make a time line:


    A: t=0, initial velocity = 0, this is the start of the entire "trip"
    B: t=very small time interval, velocity (the one the problem wants)=? This is the start of the motion

    C: t=whatever the value is, V=0. This is when the cat reaches its highest point, and starts accelerating down the y axis.
    Last edited: May 18, 2015
  7. May 18, 2015 #6
    You would (for that problem) (I think) use

    Δx = .5 m
    t = .3 s
    a = -9.8 m/s2

    Δx = ½at2 + vot

    and solve for Velocity Initial. Now obviously those numbers are made up, but I believe this would be what you are looking for, given the distance and time.

    [To the mods and others, I've been known to do too much when helping with homework, does this go too far? Thanks]
  8. May 18, 2015 #7
    I know how to solve it, my problem isn't with using the equations, my problem is more with the notation . I edited the post above, I don't know if you have read it.
  9. May 18, 2015 #8


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    Staff: Mentor

    There are two separate parts to the cat's motion. The first is while it has contact with the ground it is speeding up under muscle power. Once it loses contact with the ground its motion is dictated by gravity. In the analysis, each part of this motion can be attributed its own initial & final velocities and acceleration.
  10. May 18, 2015 #9
    There we go. I guess I misunderstood what you were asking, and no, I had not seen your edit. This is much more helpful lol.
  11. May 18, 2015 #10
    Okay, so my timeline is correct.

    Thanks all.
  12. May 18, 2015 #11


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    Staff: Mentor

    To amplify further .... the final velocity of the first part becomes what you use as the initial velocity of its flight under gravity.

    There is then a third part, somewhat more obvious, where it reconnects with the ground and uses the spring in its limbs to bring its body safely to a halt in just a few cm.
  13. May 19, 2015 #12

    rude man

    User Avatar
    Homework Helper
    Gold Member

    It says 'as it leaves the ground'. By that time it has an initial velocity which is extremely easy for you to calculate.
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