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Constant velocity physics problem

  1. Oct 14, 2007 #1
    1. The problem statement, all variables and given/known data

    When you are 20m away from your bus it begins accelerating at 3 m/s/s (from rest). With what constant velocity should you run at to catch the bus?

    2. Relevant equations

    3. The attempt at a solution
    I've done loads of ways.... and they are all wrong (spent hours). I just a want hint in the right direction as I'd like to solve this by myself. Can someone take me step by step?
    Last edited: Oct 14, 2007
  2. jcsd
  3. Oct 14, 2007 #2
    I was unable to come up with an actual answer also. Using the equation

    [tex]x_2 = x_1 + vt + \frac{1}{2}at^2[/tex]

    for the person:
    [tex]x_1[/tex] = 0
    a = 0

    so you get: [tex]x_2 = v_1t[/tex]

    for the bus:
    [tex]x_1[/tex] = 20
    [tex]v_1[/tex] = 0
    [tex]a = 3m/s^2[/tex]

    so you get: [tex]x_2 =20 + \frac{1}{2}3t^2[/tex]

    So, I ended up with two equations and three unknowns, and can only solve in terms of another variable. I tested different time values to get [tex]x_2[/tex] then solved for v of the person, and it changes from second to second, which makes sense when you think about it. So without a distance travelled or a time where they meet, I don't think there's an actual value i.e. 5 or 10 or 15m/s for velocity.
  4. Oct 14, 2007 #3
    so you're saying that theres no solution?! lol

    wow... I tried a few ways... and since V(of person) is constant, I just used that as a constant.

    Let me show you:


    Distance of Man= 20(metres)+1/2at^2=velocity x time

    Rearranged it into: 3t^2-2vt+40=0

    it's in the form ax^2+bx+c=0 So i use the [tex]x= -b\pm \sqrt{b^{2}- 4ac}}/2a[/tex]

    I get [tex]t= \frac{2v\pm 2v\sqrt{480}}{6}[/tex]

    Another way I tried was:

    (velocity of man)[tex]=\frac{20+1/2at^{2}}{t}[/tex]=at (velocity of bus)





    t=3.6515 seconds

    therefore distance bus travelled[tex]=1/2a(\sqrt{\frac{40}{3}})^{2}=20 metres[/tex]

    therefore man has to travel 20+20=40 metres

    therefore velocity of man[tex]= 40/ \sqrt{\frac{40}{3}}=10.95 m/s[/tex]

    Is the above correct??
    Last edited: Oct 14, 2007
  5. Oct 14, 2007 #4
    I think there are a couple of mistakes in your calculations.

    First with [tex] 3t^2 - 2vt + 40 = 0[/tex] using the quadratic formula you get: [tex]\frac{2v \pm \sqrt{2v^2 - 480}}{6}[/tex]
    you can't just take the [tex]2v^2[/tex] out from under the [tex]\sqrt{..}[/tex] so that leaves t as a bunch of stuff.

    I'm confused by the math of the second equation, but before that, I'm confused as to what each side is. On the left, which is the 'velocity of man' you have acceleration, but the man is traveling at at a constant velocity which means no acceleration. I'm going to look some more to see if you're trying to say something else but just wrote it wrong.

    And just because I haven't found an answer doesn't mean there isn't one. It just means I haven't found one yet.
    Last edited: Oct 14, 2007
  6. Oct 14, 2007 #5

    Yes the quadratic formula bit was wrong.

    As for the 2nd method, what im trying to say is... the velocity=distance divided by time right?

    Okay... so the distance the Man has to travel is 20 metres in addition to the distance the bus travelled in that time i.e. 1/2at^2 and the usage of the acceleration shouldn't make a difference cos acceleration is metres/second/second, and your timesing it by t^2 therefore you have metres as units.

    The RHS of the equation i.e. at is the distance travelled by the bus (as theres no initial velocity=0) so that should be correct i think.
  7. Oct 14, 2007 #6
    Your answers work for the original problem, but when I did it there was no one answer. For example, if t = 2s, v = 13m/s, that works, as well as t=3s, v=11.167m/s. I think without a definite time or velocity, the answer is some sort of curve.

    And you're equating velocity of the person to velocity of the bus, but there's no saying that they are or should be equal. Or, you may be using distance of the person to be equal to velocity of the bus, and I don't think that's right either. It's still not clear to me. But let me say, I'm not an expert, just another person learning and trying to help as I go.
  8. Oct 15, 2007 #7
    Anyone else??
  9. Oct 15, 2007 #8
    What chocokat did looks correct, you have some info misssing (probably the question wants minimum velocity, which you can find).
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