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Non linear motion ( Physics olympiads question )

  1. Jul 31, 2010 #1
    1. The problem statement, all variables and given/known data

    A cat sitting in a field suddenly sees a standing dog. To save its life, the cat runs away in a straight line with speed u. Without any delay, the dog starts with running with constant speed v>u to catch the cat. Initially, v is perpendicular to u and L is the initial seperation between the two. If the dog always changes its direction so that it is always heading directly at the cat, find the time the dog takes to catch the cat in terms of v, u and L.

    2. Relevant equations
    As of the time of posting this. I'm not sure if there's a simple elegant solution to this, but the history of olympiads questions tells me it's probably a really complicated solution.
    It's probably more logic and calculus than any kinematics equation.
    3. The attempt at a solution

    So far i've gotten to resolving U as a vector into it's horizontal and vertical component. Taking the instantaneous direction of V as the horizontal. Taking the angle between U and the horizontal as A.
    This yields the velocity of the dog, relative to the cat as V - UcosA
    I'm thinking of graphing V-UcosA with time. Not really sure how to proceed from here. '
     
    Last edited: Jul 31, 2010
  2. jcsd
  3. Jul 31, 2010 #2
    You are right about the relative velocity [tex]v-u \cos \alpha[/tex]. Now, there is some calculus involved. Let [tex]s[/tex] be the distance between the cat and the dog at an arbitrary moment. So we have

    [tex]\frac{-ds}{dt}=v-u \cos \alpha[/tex]

    [tex]-\int_{L}^{0} ds=\int_{0}^{T} (v-u \cos\alpha) dt=L[/tex] (1)

    [tex]T[/tex] is the sought time.

    Since both of them travel the same distance in x-direction, we have the second condition

    [tex]uT=\int_{0}^{T} v\cos\alpha dt[/tex] (2)

    Solving (1) i (2) yields

    [tex]T=\frac{uL}{v^{2}-u^{2}}[/tex].

    I don't know if there is any simple solution (not involving calculus).
     
  4. Jul 31, 2010 #3
    N-Gin, how do you solve (1) and (2)?
    P.S: There is indeed at least one non-calculus solution.
     
  5. Jul 31, 2010 #4
    Yes i was about to ask as well. How did you solve (1) and (2)?

    And hikaru would you kindly enlighten us as to the method/solution then? I'm trying to work it out but it has been hours. I'm really burnt over this question so a direct answer would be appreciated =)

    P.S the answer is incorrect. I think it might be because the speed velocities become a constant after awhile and it isnt accounted for in the equations involving cosA. ( After the vertical component is covered the velocity only goes horizontally and so it becomes a constant. while VcosA would just continue in it's sinusoidal curve. therefore giving an inaccurate area under the graph )
     
  6. Jul 31, 2010 #5
    @N-Gin: It should be [tex]T=\frac{vL}{v^2-u^2}[/tex]
    @leucocyte: The simple solution lies in N-Gin's equations! It's quite fun, so you should try thinking. Just one more step and you will see the answer. Hint: There is something about subtraction or addition :rolleyes:

    The dog's velocity never becomes a constant. The problem is limited from the moment the dog sees the cat to the moment the dog catches the cat, so there is no periodic thing.
     
  7. Jul 31, 2010 #6
    I'd like to clean my post up a bit. I was thinking while writing.

    First, we take the direction the cat runs as the horizontal axis. The cat runs in a straight line and so it's vertical position is constant. We take the dog as sitting at the origin and hence the cat must be at the position (0,L) and the dog at position (0,0) initially.

    Writing the position of the cat and dog as two vectors [tex]\vec{c}[/tex], and [tex]\vec{d}[/tex], we require the distance between the two critters to be 0, that is [tex]||\vec{c} - \vec{d}||[/tex] must be 0.

    We consider the identity [tex]||\vec{c} - \vec{d}||^2 = ||\vec{c}||^2 - 2||\vec{c}|| ||\vec{d}||\cos{\beta} + ||\vec{d}||^2[/tex], where [tex]\beta[/tex] represents the angle between the vectors c and d. At the time when the two animals meet, the angle between the two vectors, [tex]\beta[/tex] is 0 which gives an equation for when the distance between the two animals is 0,

    [tex] ||\vec{c}||^2 + ||\vec{d}||^2 = 2||\vec{c}|| ||\vec{d}||[/tex].

    Writing [tex]\vec{c} = (ut, L)[/tex] and [tex]\vec{d} = (v_xt,v_yt)[/tex] and noticing that [tex]||\vec{d}|| = ||\vec{v}||t[/tex], we find that the two animals meet when,

    [tex] (uT)^2 + L^2 + (||\vec{v}||T)^2 = 2||\vec{v}T||^2[/tex],

    where I exploit the fact that when they meet, the two vectors will be the same distance away from the origin, i.e. [tex]||\vec{c}|| = ||\vec{d}||[/tex].

    Finally,

    [tex]T^2(||\vec{v}||^2 - u^2) = L^2[/tex],

    or,

    [tex] T = \frac{L}{\sqrt{||\vec{v}||^2 - u^2}}[/tex].

    I do realize that this may be horribly wrong. Please be gentle.
     
    Last edited: Jul 31, 2010
  8. Jul 31, 2010 #7

    kuruman

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    It is best to consider the motion transverse to the cat's velocity. You know that the dog covers distance L in that direction. If we let α be the angle between the dog's velocity and the direction of L, then

    [tex]L=\int^{T}_{0}v \ Cos \alpha(t) \ dt[/tex]

    But

    [tex]Cos \alpha(t)=\frac{L}{\sqrt{L^2+u^2+t^2}}[/tex]

    Therefore,

    [tex]1=v\int^{T}_{0}\frac{dt}{\sqrt{L^2+u^2t^2}}[/tex]

    When you do the integral and solve for T, you get a very neat and compact answer. I will not spoil your fun by telling you what it is.
     
  9. Jul 31, 2010 #8
    @kuruman:
    Hrm. This seems to have some problems with it as well. In particular we should expect that if the dog and cat are running at the same speeds (i.e. ||v|| = ||u||), that the dog will never catch the cat. The solution found from that integral doesn't seem to reflect this.

    @hikaru:
    This also seems to neglect some intuition, in particular for the case where the cat is faster than the dog. Unfortunately I wasn't able to follow you or N-Gin :( on the derivation of that formula and so I'm not really sure. N-Gin seems to have done some magical things with those integrals considering that the angle is dependent on time.
     
    Last edited: Jul 31, 2010
  10. Jul 31, 2010 #9

    kuruman

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    Of course in that case the dog will never catch the cat, but you are changing the problem which clearly states that the dog's speed is greater than the cat's (v>u).

    *** On edit ***
    Let me think about this some more.
     
    Last edited: Jul 31, 2010
  11. Jul 31, 2010 #10
    It is changing the question, however your answer should be able to reflect the fact that there is no positive time for which the dog will reach the cat. The derivation of the solution remains the same regardless of whether the dog is faster.
     
  12. Jul 31, 2010 #11

    kuruman

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    You are correct. The time that I indicated, is Ty, i.e. what is required for the dog to reach distance L vertically, but it does not guarantee that it will be at the same position as the cat horizontally. I thought that pointing the dog in the cat's direction did that. So, I guess, one has to do a similar calculation in the horizontal direction to find time Tx and require that Tx=Ty. I don't see a more elegant solution at present, but I will keep thinking about it.
     
  13. Jul 31, 2010 #12
    Hi thanks for the help so far everyone, i often browse through physics forums but only felt compelled to register after this question.

    Kuruman could you help me to understand how you derived that equation? I follow the rest of your workings. And i think with the addition of equating the 2 components you would get the right answer.

    Coto, i'm not sure if you're right the answer given is not worked in vectors. But of course with most questions like that there is more than one way to work towards the answer. I do not see anything wrong with your workings, unless other PF members can spot a problem I would venture to say you've gotten it right
     
  14. Jul 31, 2010 #13

    vela

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    Coto's solution isn't correct because the location of the dog isn't given by (vxt, vyt). The direction of the dog's velocity changes with time as it follows the cat.
     
  15. Jul 31, 2010 #14
    I've rechecked the solution N-Gin is actually right provided that it is ( as hikaru says ) VL not UL, but I'm still a little confused as to how he did the last step?
     
  16. Jul 31, 2010 #15

    vela

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    Hint: You can write equation (1) as

    [tex]vT - u \int_0^T \cos \alpha \, dt = L[/tex]
     
  17. Jul 31, 2010 #16
    Ahh yes that makes much more sense now. The morning coffee probably helped as well. Thanks all for responding so quickly to this question =)
     
  18. Jul 31, 2010 #17

    kuruman

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    My mistake. I meant to write

    [tex]
    Cos \alpha(t)=\frac{L}{\sqrt{L^2+u^2t^2}}
    [/tex]

    One plus sign too many. However, I am not convinced my method is correct. Angle alpha as defined above is the angle between the position vector and the L-axis. The instantaneous velocity is not in the same direction as the position vector, so alpha should not be used to find the components of the velocity.
     
  19. Jul 31, 2010 #18
    @Coto: [tex]\vec{d}\neq (v_xt;v_yt)[/tex] and [tex]|\vec{d}|\neq |\vec{v}|t[/tex] as the dog's velocity changes its direction with time.
    @leucocyte: I guess you have found the non-calculus solution?
     
  20. Jul 31, 2010 #19
    @hikaru: Not quite yet, but i have 3 essays to rush out and some other tutorials to finish so im gonna leave looking for the non-calculus solution aside for now unless you would care to kindly enlighten me? =)
     
  21. Aug 1, 2010 #20
    Okay. Here are the 2 equations by N-Gin:
    (1): [tex]\int ^T _0 (v-ucosA)dt=L[/tex]
    (2): [tex]uT=\int ^T _0 vcosAdt[/tex]

    Change them a little bit:
    (1): [tex]vT-u\int ^T _0 cosAdt = L[/tex]
    (2): [tex]uT/v = \int ^T _0 cosAdt[/tex]

    Now substitute [tex]\int ^T _0 cosAdt[/tex] from (2) to (1), and solve for T :wink:
     
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