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Thread: shortest distance

  1. #1
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    shortest distance

    At a certain instant ship Q is at a distance of 4a due east of ship P. Q is moving northwards with constant speed u and P is travelling with constant speed 2u. Find the direction of P if it is to intercept Q.Find the time T, in terms of a and u, it would take P to intercept Q.If, instead, after time has elapsed, the speed of P drops to constant speed u, without changing direction, find, in terms of a the shortest distance between P and Q the distance each ship has moved from its original position to its position when they are closest together.distance in j direction should be equal
    2usina*t=ut
    a=30 degrees.
    distance in x direction =4a
    2ucosa*T=4a
    T=4a/u√3 after T/2 it has moved 2ucos30(T/2) =2a east bit it has also moved up north. normally when i find the shortest distance between the ships they are horizontal with each other but here both ships have moved up different distances north so I can not.
    Last edited by markosheehan; Jun 27th 2017 at 09:48 AM.
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    Re: shortest distance

    Your post is too cluttered for me...
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    Re: shortest distance

    The $x$ component of distance:

    $\Delta x = 4a-\left[2u (\cos \theta) T_1 + u(\cos \theta) (t-T_1)\right]$

    The $y$ component of distance:

    $\Delta y = ut - \left[2u (\sin \theta) T_1 + u(\sin \theta) (t-T_1)\right]$

    The magnitude of the distance:

    $s = \sqrt{\left(4a-u(\cos \theta) T_1 - u(\cos \theta )t\right)^2+(ut-u(\sin \theta) T_1 - u(\sin \theta) t)^2}$

    $\dfrac{ds}{dt} = \dfrac{-u(\cos \theta) (4a-u(\cos \theta) (T_1+t))+(u-u\sin \theta)(ut-u(\sin \theta) (T_1+t))}{\sqrt{\left(4a-u(\cos \theta) T_1 - u(\cos \theta) t\right)^2+(ut-u(\sin \theta) T_1 - u(\sin \theta) t)^2}}$
    Last edited by SlipEternal; Jun 27th 2017 at 02:01 PM.
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  4. #4
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    Re: shortest distance

    Quote Originally Posted by markosheehan View Post
    At a certain instant ship Q is at a distance of 4a due east of ship P. Q is moving northwards with constant speed u and P is travelling with constant speed 2u.
    So, the position of Q relative to the initial position of P, at time t after the initial time, is <4a, ut>. Taking the angle, clockwise from N, of the velocity vector of P, to be theta, P's position will be <2u cos(t), 2u sin(t)>.
    (You say "constant speed". I assume that is constant velocity- that is, that P is moving in a straight line.)

    Find the direction of P if it is to intercept Q.
    P will intercept Q at time t, if and only if <4a, ut>= <2ut cos(theta),2ut sin(theta)>. From ut= 2ut sin(theta), sin(theta)= 1/2.

    Find the time T, in terms of a and u, it would take P to intercept Q. If, instead, after time has elapsed, the speed of P drops to constant speed u, without changing direction, find, in terms of a the shortest distance between P and Q the distance each ship has moved from its original position to its position when they are closest together.distance in j direction should be equal
    2usina*t=ut
    a=30 degrees.
    Yes. However, since you were given the initial distance apart to be "4a", you should not use "a" again to mean the angle.

    distance in x direction =4a
    2ucosa*T=4a
    T=4a/u√3 after T/2 it has moved 2ucos30(T/2) =2a east bit it has also moved up north. normally when i find the shortest distance between the ships they are horizontal with each other but here both ships have moved up different distances north so I can not.
    You must have both north-south and east-west equations satisfied. You have already found the angle so that the north-south equation is satisfied for all t (you were able to get "sin(theta)= 1/2" so "theta= 30 degrees" because u and t in "2usina*t=ut" cancel ) so it is only a matter of satisfying the east-west equation.
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  5. #5
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    Re: shortest distance

    Quote Originally Posted by SlipEternal View Post
    The $x$ component of distance:

    $\Delta x = 4a-\left[2u (\cos \theta) T_1 + u(\cos \theta) (t-T_1)\right]$

    The $y$ component of distance:

    $\Delta y = ut - \left[2u (\sin \theta) T_1 + u(\sin \theta) (t-T_1)\right]$

    The magnitude of the distance:

    $s = \sqrt{\left(4a-u(\cos \theta) T_1 - u(\cos \theta )t\right)^2+(ut-u(\sin \theta) T_1 - u(\sin \theta) t)^2}$

    $\dfrac{ds}{dt} = \dfrac{-u(\cos \theta) (4a-u(\cos \theta) (T_1+t))+(u-u\sin \theta)(ut-u(\sin \theta) (T_1+t))}{\sqrt{\left(4a-u(\cos \theta) T_1 - u(\cos \theta) t\right)^2+(ut-u(\sin \theta) T_1 - u(\sin \theta) t)^2}}$
    Sorry, I had to leave work. Here, $T_1$ is the time at which ship P changes speed from 2u to u. Note that inertia would prevent an instantaneous change of speed like this, but unless the problem stated that P gradually decelerated from a speed of 2u to u, you don't have to worry about that, and it will not affect the model by much.

    Next, set the derivative equal to zero and solve for $t$.

    And as both you and HallsofIvy found, $\theta = 30^\circ$.

    This gives:

    $0 = -(\cos \theta) (4a-u(\cos \theta) (T_1+t))+(1-\sin \theta)(ut-u(\sin \theta) (T_1+t))$

    $0 = u(cos^2 \theta)(T_1+t)-4a\cos \theta + (ut-u(\sin \theta)(T_1+t)-ut\sin \theta + u(\sin^2\theta)(T_1+t))$

    $0 = u(T_1+t)-4a\cos \theta + u(t-(T_1+t)\sin \theta-t\sin\theta)$

    $0 = 2uT_1-uT_1 - 4a\sqrt{3} +4ut -2ut$

    $0 = uT_1-4a\sqrt{3}+2ut$

    $t = \dfrac{4a\sqrt{3}-uT_1}{2u}$

    You may want to check my work. It is entirely possible I made an algebra mistake.
    Last edited by SlipEternal; Jun 27th 2017 at 04:34 PM.
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