## Vectors question.

Hello,

I'd like to please verify whether my solution to the problem as described below is indeed correct.

Ship A is traveling east at a constant velocity of 30km/h. It notices ship B 15 km away and 30° from the north. Ship B is traveling at a constant velocity. 20 minutes later ship A notices ship B 45° from the north, and 20 more minutes after that ship B is observed 60° from the north. The distance between ship A and B at that point in time ought to be calculated, including the velocity (magnitude and direction) of ship B.

Proposed solution:

Supposing ship A first spots ship B from O, (0,0):

1) (1/3Va)(OB + 1/3Vb) / |(1/3Va)||(OB + 1/3Vb)| = 1/SQRT(2)

2) (2/3Va)(OB + 2/3Vb) / |(2/3Va)||(OB + 2/3Vb)| = SQRT(3)/2

[Va, Vb, OB are vectors; 1/SQRT(2) = cos 45°; SQRT(3)/2 = cos 30°]

I got Vb = (27.56,15.91), |Vb| = 31.82 km/h, distance between ship A and B 40 minutes later = 12.14 km

Are these equations correct? Is their solution?

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 Does 30 degrees from north mean 30 degrees from north towards east or 30 degrees from north towards west?

Recognitions:
Homework Help
 Quote by peripatein 1) (1/3Va)(OB + 1/3Vb) / |(1/3Va)||(OB + 1/3Vb)| = 1/SQRT(2)
Can you explain the logic behind that equation? It appears to be saying that angle A'OB' is 45 degrees, where A' and B' are the positions after 20 minutes.

## Vectors question.

30 degrees from the north means 60 degrees from the east. The movement is only in the east-west plane. That is my assumption from the details in any cass.

 I think there might be two solutions but if we assume 30 degrees from north means towards east I think your numbers are off. Using ruler and compass I solved the problem graphically. I was careful and think my number are off less then 10%. See below, Attached Thumbnails
 Recognitions: Homework Help Science Advisor As I indicated before, I don't think your equations are right. Your equation (1) appears to be saying that angle A'OB' is 45 degrees, where A' and B' are the positions after 20 minutes. It's the angle between OA' and A'B' that's 45 (or 135) degrees.
 Hi, For the problem as described above, are the following two equations correct? Supposing A = (Va,0)t and B = (15cos60,15sin60) + (Vbx, Vby)t Thus: 20 minutes later: A = (1/3Va,0) and B = (15cos60,15sin60)+(1/3Vbx,1/3Vby) 40 minutes later: A = (2/3Va,0) and B = (15cos60,15sin60)+(2/3Vbx,2/3Vby) Hence: (1) 1/SQRT(2) = (1/3Va,0)*(15cos60+1/3Vbx-1/3Va,15sin60+1/3Vby) / |(1/3Va,0)||(15cos60+1/3Vbx-1/3Va,15sin60+1/3Vby)| (2) SQRT(3)/2 = (2/3Va,0)*(15cos60+2/3Vbx-2/3Va,15sin60+2/3Vby) / |(2/3Va,0)||(15cos60+2/3Vbx-2/3Va,15sin60+2/3Vby)| Is that correct?
 In case these last equations are also wrong, please suggest any other method/set of equations by which this problem may be approached and solved.
 Recognitions: Homework Help Science Advisor Let's approach this geometrically. Let A, B be the initial positions, A', B' the positions after 20 minutes, and A", B" after 40. Spinnor's diagram is useful. What angle are we told is 45 degrees? What lines border that angle? What vectors (treating A as the origin) correspond to the orientations of those lines? Your equation (1) appears to be using OA', OB' as those vectors.
 I have made some corrections to my initial equations. They are valid now, wouldn't you agree?

 Quote by peripatein .... (1) 1/SQRT(2) = (1/3Va,0)*(15cos60+1/3Vbx-1/3Va,15sin60+1/3Vby) / |(1/3Va,0)||(15cos60+1/3Vbx-1/3Va,15sin60+1/3Vby)| (2) SQRT(3)/2 = (2/3Va,0)*(15cos60+2/3Vbx-2/3Va,15sin60+2/3Vby) / |(2/3Va,0)||(15cos60+2/3Vbx-2/3Va,15sin60+2/3Vby)| Is that correct?
(1) 1/SQRT(2) = (1/3Va,0)*(15cos60+1/3Vbx-1/3Va,15sin60+1/3Vby) / |(1/3Va,0)||(15cos60+1/3Vbx-1/3Va,15sin60+1/3Vby)|

simplify -->

(1) 1/SQRT(2) = (15cos60+1/3Vbx-1/3Va) / |(15cos60+1/3Vbx-1/3Va,15sin60+1/3Vby)|

Do the same for the other eq.

 The equations indeed seem correct, but I have not been able to extricate Vbx and Vby. May you please try it yourself and suggest how to go about it?
 Recognitions: Homework Help Science Advisor I would start by making the equations simpler to write out. Put x for Vbx etc., multiply out to get rid of the thirds, maybe turn cos 60 and sin 60 into numbers. You have two quadratic equations with two unknowns. In general, that could give you a quartic, but maybe you get lucky. Post what you do get.
 Hi, It is always possible that I have made some mistakes along the way, but I have got: (1) x^2-37.5x -67.5SQRT(3)y-3y^2-787.5=0 (2) x^2-15x+45SQRT(3)y-y^2-1462.5=0 Which doesn't get me anywhere. Any advice?
 I did make a mistake, and will repost the correct equations shortly
 Hi, I still get: (1) x^2 -37.5x -67.5SQRT(3)y -3y^2 -787.5=0 (2) x^2 -15x -45SQRT(3)y -y^2 -1462.5=0 Which doesn't get me anywhere. Any advice?
 Recognitions: Homework Help Science Advisor If you take the difference of the two equations that will remove the x^2 term, leaving an expression for x in terms of y and y^2. You can use that to substitute for x in one equation. That produces a quartic in y (as I feared). You can look up how to solve quartics. I'm surprised the problem is this hard, but the logic all seems correct.