Tips for solving word problems of vector?

I will be taking PreCalc next school year. I have prepared for it by using Khan Academy. Now, I am learning vectors. I can calculate the sum and difference of vectors. However, I am not good at solving word problems of vectors. Could you give me some tips for solving it well?

1 Answer

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  • Sean
    Lv 5
    1 month ago

    An example would be helpful oh well let's make something up

    Two friends A and B go camping.  One morning A goes out for a walk with speed of 4K/Hr while B whose turn it is does the dishes. A who has bad handwriting leaves a message saying he will be travelling north for an hour and then west for three quarters of an hour and invites B to meet him there.  B mixes up the order directions and times and sets out fifteen minutes later not realizing that his walking speed is only 3K/Hr after 75 minutes he realizes his error and calls A on a cell phone. A advises B to stop and said that he would join him where B is

    i) where is  A when B calls relative to the campsite

    ii)where is B when the call is made relative to the campsite

    iii) how far must A travel and in what direction

    i) At the time of call one and one half hours (75 minutes and the quarter hour head start) have passed for A who has gone one hour north and only a half hour west

    this is a right triangle 1*4 kilometres and 1/2* 4 kilometres west

    changing to vector form with campsite at (0,0) the position is (-2,4) so

    cos(alpha) =-2/sqrt((-2)^2 + 4^2)) = -2/sqrt(20) = -1/sqrt(5) = 116.6 degrees from the east relative to the campsite and sqrt(20) kilometres 4.5 kilometres away

    ii) B has gone one hour west and one quarter hour north at 3K/hr

    so the position is (-3*1,3/4)  so cos(beta) = -3/sqrt((-3)^2 + (3/4)^))

    166.0 degrees from the campsite measuring from due east a distance of

    sqrt(9 + 9/16) = sqrt(153/16) = 3.09 kilometres

    iii)  A has reached the position (-2,4) and B (-3,3/4) to get the direction A must travel subtract A's vector (from which he sets out) from B's vector

    (-3 - -2,4-3/4) = (-1,-13/4) which is cos(gamma) = -1/sqrt((-1)^2 +(13/4)^2)) = -1/sqrt(185/16) for a distance of 3.4 kilometres at a bearing of 252.9 degrees from due east

    NOTES for practical problems converting vectors into an (a,b) form is usually the easiest way to proceed for example a bearing southeast for 20 kilometres is

    20(cos theta, sin theta) since southeast is the fourth quadrant gives 20 (cos (305-270), sin(305-270))

    assuming angles are measured counterclockwise the formula is

    0 <= theta <90 convert theta to theta - 0

    180 >theta>= 90 theta converts to  theta -90

    270> theta >=180  theta converts to theta - 180

    360 > theta >=270 theta converts to theta - 270

    if coordinates (a,b) are turned into k(cos(x),sin(x))

    calculate

    "cos" as |a|/sqrt(a^2 + b^2)

    "sin" as |b|/sqrt(a^2 + b^2)

    "tan" as |b|/|a|

    a>=0,b>=0  arccos("cos"),arcsin("sin"),arctan("tan")

    a < 0 b>= 0 180-arccos("cos"),180-arcsin("sin"),180-arctan("tan")

    a<0 b<=0     180 + arccos("cos"), 180 + arcsin("sin"), 180 + arctan("tan")

    a>0,b<=0    360 - arccos("cos"),360-arcsin("sin"),360-arctan("tan")

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