A cruise ship leaving port travels 52.0 km 45.0° north of west and then 65.0 km

Question

A cruise ship leaving port travels 52.0 km 45.0° north of west and then 65.0 km at a heading 30.0° north of east. (a) Find the ship's displacement vector. 67.2 R, Find the components of each displacement, then add = components. For instance, the sum of the two x components is the x component of the resultant. km 49 R, Find the components of each displacement, then add -components. For instance, the sum of the two x components is the x component of the resultant. km (b) Find the displacement vector's magnitude and direction. Magnitude Direction km o north of east Need Help? Talk to e Tuter

Explanation / Answer

Solution:-

Rx = -52*cos(45) + 65*cos(30) = 19.52km (note the first vector point to the left while the second vector points to the right)

Ry = 52*sin(45) + 65*sin(30) = 69.27km

Therefore the magnitude = sqrt(19.52^2 + 69.27^2) = 71.97km

= arctan(69.27/19.52) = 74.26o

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