• AravindG
10. The smoke emitted from a steamer, sailing due south, extends in a direction making an angle of 30⁰, north of west, while that from another steamer, sailing due north, with same speed, extends in a direction making an angle Of 30⁰, west of south. If v be the speeds of each steamer, find the true velocity of wind.
  • Stacey Warren - Expert
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  • jamiebookeater
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  • TuringTest
Here's what I got...|dw:1327860852617:dw|V1 and V2 are the velocities of the first and second ships respectively. S1 and S2 are the velocities of the smoke. The way I see it we have a system of equations based on these vectors if we notice that the smokes vectors are the sum of the ship and wind vectors.\[S_{1}=V_1+W\to W=S_1-V_1\]\[S_2=V_2+W\to W=S_2-V_2\]the magnitude on each speed vector the the boats is the same, so we have\[S_1-V_1=<|S_1|\cos\theta_1,|S_1|\sin\theta_1>-<0,-|V|>\]\[=<|S_1|\cos\theta_1,|S_1|\sin\theta_1+|V|>\]\[S_2-V_2=<|S_2|\cos\theta_2,|S_2|\sin\theta_2>-<0,|V|>\]\[=<|S_2|\cos\theta_2,|S_2|\sin\theta_2-|V|>\]these are equal quantities (both equal to wind) so by treating their components separately we can create the following system\[|S_1|\cos\theta_1-|S_2|\cos\theta_2=0\]\[|S_1|\sin\theta_1-|S_2|\sin\theta_2=-2|V|\]This system can be solved for S1 and S2 in terms of V. Once you have that you can use either of the earlier equations for W to find it in terms of the speed of the boat,
  • TuringTest are what I mean by theta 1 and 2 just to clarify:|dw:1327862701263:dw|

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