q1.What is the distance of the fourth leg? q2What is the value of Θ4? q3.What is the value of Dy?
A sailboat race course consists of four legs defined by the displacement vectors A , B , C and D shown above.
The values of the angles are Θ 1 = 35 0 , Θ 2 = 34 0 , and Θ 3 = 20 0 .
The magnitudes of the first three vectors are A = 3 km, B = 5 km and C = 4.5 km. The finish line of the course coincides with the starting line.
The coordinate system for this problem has positive x to the right, positive y as up and counter-clockwise to be a positive angle.
Image : https://www.flipitphysics.com/Content/smartPhysics/Media/UserData/ebfae213-ad62-da58-2e01-ff6f3e3c551b/sail_boat.gif
1 Expert Answer
Jeremie F. answered • 06/02/23
PhD in Physics with 5+ years of teaching experience.
If you keep track of the total displacement in x and y, this problem becomes simple as it just will need pythagorean theorem at the end. So let's see how this works.
Note that in the first leg (A) the vector points in the positive x and positive y directions, so as a sanity check we should find (if we did it correctly) that x and y are positive.
Note that the angle is 35 degrees (when using the calculator make sure it is in degree mode!).
Now we need some trigonometry, note SOH, CAH, TOA (you may need to review this if it does not look familiar to you). The x component of the vector can be calculated by using CAH, which means that
cos(35) = A/H,
but H is the hypotenuse and we know the length of H. H = 3km.
cos(35) = A/3 which means 3cos(35) = A
note that A in this case is just the x component so
3cos(35) = x and thus x = 2.4575 km (rounded to 4 decimal places).
Similarly for y
sin(35) = O/H which means sin(35) = O/3 and finally
3sin(35) = y (since O is the y direction in this triangle).
y = 1.7207 km (again rounded to 4 decimals)
Now for the second leg
For our sanity check, note that this vector is pointing in the NEGATIVE x direction and the positive y direction. So when we do our calculations this is what we should find.
Note that from the diagram theta_2 is the angle measured away from 180 degrees in this case so the angle that we need is 180 - theta_2 which is 180-34 = 146 degrees.
This leg was 5 km long so (again by making the same triangles we did before)
x = 5cos(146) = -4.1452 km (rounded to 4 decimals, I will just round all to 4 decimals from here on out)
y = 5sin(146) = 2.7960 km
Now for the third leg. Note that this points in the negative x and negative y direction so we should find they are both negative.
Note here the angle is again measure from 180 degrees, but this time the angle should be greater than 180 (look at the angle and this should become clear).
So the angle we need is 180+20 = 200 degrees.
x = 4.5cos(200) = -4.2286 km
y = 4.5sin(200) = -1.5391km
Ok now we want the total displacement away from the starting point, so just add up all of the x values and y values
x = 2.4575 -4.1452 - 4.2286 = -5.9163 km (displacement)
y = 1.7207 + 2.7960 - 1.5391 = 2.9776 km
This makes sense because we need to go in the negative y direction to get back to the starting spot, so our displacement in the y direction must have been positive!
Now we just need pythagorean theorem to figure out the length of the last leg
x^2+y^2 = D^2, so D = sqrt(x^2+y^2) = 6.6233 km.
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A sailboat race course consists of four legs, defined by the displacement vectors A, B, C and D as the drawing indicates. The magnitudes of the first three vectors are A=3.20km, B=5.10km , and C=4.80km. The finish line of the course coincides with the starting line. Using the data in the drawing, find the distance of the fourth leg and the angle θ. [6]
Expert verified solution, super gauth ai.
Explanation
Answer: since starting ending point are same A+B+C+D=0 Horizontal Component. 3: 20cos (40)-5.10cos (35)-480cos (23)+0cos 0=0 2.45-4.17-4.41+Dcos θ =0 -6.13+Dcos θ =0 Dcos θ =6.13 ①① veaurical Component. 3. 20sin (40)-5.10sin (35)-4.80sin (23)+Dsin θ =0 2.05-2.92-1.87+Dsin θ =0 Dsin θ =2.74 ② frac D'sin θ =frac 2.0= (2.74)/6.13 , tan θ =0.44, θ =23.74° Dcos 23.74°=6.13 D= (6.13)/0.91 , D=6.73 Hence, θ =23.74° 8 vector D=6.73. Explanation: Hence θ =23.74° 8 vector D=6.73
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Find the distance and angle of the fourth leg of a sailboat race course defined by four displacement vectors. See the drawing, the magnitudes of the vectors, and the solution steps.
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A sailboat race course consists of four legs, defined by the displacement vectors A, B, C, and D. Find the distance of the fourth leg and the angle θ using trigonometry.
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