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Math Help - relative velocity and interception

  1. #1
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    relative velocity and interception

    Ship A sails with velocity (30i+40j) km/h and ship B sails with velocity (15i+12j) km/h . At one instant , the position vector of ship B relative to ship A is (5i+12j) km . Calculate the magnitude and direction of the velocity of ship A relative to ship B . Calculate also the shortest distance between the 2 ships .

    My answers :

    I found the magnitude of V of A relative to B - 31.76 km/h with a direction of N28.18 E

    and the shortest distance - 1.26 km

    Am i correct ?

    This is the continuation of the question . At the instant when ship A is east of ship B , ship A changes its course but maintain its speed in order to intercept ship B . Find the course that ship A must steer and the time taken to intercept ship B

    my answers :

    course of A - 283.89 degrees .

    time - i am not really sure . I don know the distance between A and B here .
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  2. #2
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    Hello hooke
    Quote Originally Posted by hooke View Post
    Ship A sails with velocity (30i+40j) km/h and ship B sails with velocity (15i+12j) km/h . At one instant , the position vector of ship B relative to ship A is (5i+12j) km . Calculate the magnitude and direction of the velocity of ship A relative to ship B . Calculate also the shortest distance between the 2 ships .

    My answers :

    I found the magnitude of V of A relative to B - 31.76 km/h with a direction of N28.18 E

    and the shortest distance - 1.26 km

    Am i correct ?

    This is the continuation of the question . At the instant when ship A is east of ship B , ship A changes its course but maintain its speed in order to intercept ship B . Find the course that ship A must steer and the time taken to intercept ship B

    my answers :

    course of A - 283.89 degrees .

    time - i am not really sure . I don know the distance between A and B here .
    I agree with all your answers. Good work!

    In the first stage, the velocity of A relative to B is 15\textbf i + 28\textbf j, and the initial displacement of A from B is -5\textbf i -12\textbf j.

    When A is due East of B, its displacement from B is r\textbf i + 0\textbf j, for some r. Considering the change in the \textbf j component, the time taken for this is \frac{12}{28} = \frac{3}{7} hours. In this time, the increase in the \textbf i component of A's position relative to B is \frac{3}{7}\times15=\frac{45}{7}. So its position relative to B is now \left(-5+\frac{45}{7}\right)\textbf i= \frac{10}{7}\textbf i.

    In the interception phase, the velocity of A relative to B is (according to my working, and I think to yours) -48.54\textbf i. Therefore the time A takes to intercept B is:
    \frac{10}{7\times 48.54} hours = 1.77 minutes
    But check my working, won't you?

    Grandad
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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello hookeI agree with all your answers. Good work!

    In the first stage, the velocity of A relative to B is 15\textbf i + 28\textbf j, and the initial displacement of A from B is -5\textbf i -12\textbf j.

    When A is due East of B, its displacement from B is r\textbf i + 0\textbf j, for some r. Considering the change in the \textbf j component, the time taken for this is \frac{12}{28} = \frac{3}{7} hours. In this time, the increase in the \textbf i component of A's position relative to B is \frac{3}{7}\times15=\frac{45}{7}. So its position relative to B is now \left(-5+\frac{45}{7}\right)\textbf i= \frac{10}{7}\textbf i.

    In the interception phase, the velocity of A relative to B is (according to my working, and I think to yours) -48.54\textbf i. Therefore the time A takes to intercept B is:
    \frac{10}{7\times 48.54} hours = 1.77 minutes
    But check my working, won't you?

    Grandad
    hello grandad , i got 2.58 minutes . Err , could you check my working for my mistake ?

    The ship move to A' when its exactly east of B , and since the course changes , Va=xi+yj

    velocity of A relative to B= Va-Vb = xi+yj-(15i+12j)

    b-12=0 , b=12 (vector j is 0)

    |Va|=50

    \sqrt{a^2+b^2}=50

    a=48.54

    so Va=48.54i+12j

    so direction would be N 76.1 E

    the shortest distance is 1.26 km , sin 5.56=d'/13 ,d'=1.43

    aVb=33.54

    s=vt , 1.43=33.54t , t=0.043 hours = 2.58 mins .
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  4. #4
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    Hello hooke
    Quote Originally Posted by hooke View Post
    hello grandad , i got 2.58 minutes . Err , could you check my working for my mistake ?

    The ship move to A' when its exactly east of B , and since the course changes , Va=xi+yj

    velocity of A relative to B= Va-Vb = xi+yj-(15i+12j)

    b-12=0 , b=12 (vector j is 0)

    |Va|=50

    \sqrt{a^2+b^2}=50

    a=48.54

    so Va=48.54i+12j

    so direction would be N 76.1 E

    the shortest distance is 1.26 km , sin 5.56=d'/13 ,d'=1.43

    aVb=33.54

    s=vt , 1.43=33.54t , t=0.043 hours = 2.58 mins .
    The shortest distance, 1.26 km, has nothing to do with it. Ship A changes course when it is due East of B, not when it's at the shortest distance from B.

    When A is due East of B, then, as I showed before, it is \tfrac{10}{7}= 1.429 km away, and its velocity relative to B is then -48.54\textbf i (which you have confirmed in your working). So this distance 1.429 km is being 'closed down' at a rate of 48.54 km per hour. So the time A takes to intercept B is
    \frac{1.429}{48.54}=0.0294 hours
    =1.77 minutes
    Grandad
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