Displacement Vector of Ship: Find & Calculate

[bmchenry]

Homework Statement

a radar operator of a cutter observes that at 10h20m a ship is at a distance of 9.5km on a bearing of 60defrees East of North and at 11h10m it is at a distance of 4.2km on a bearing of 33 degrees East of North

a. Measured from its postion at 10h30m what is the displacement vector of the ship at 11h10m
b. assuming the ship continues on the same course at the same speed what will be its displacement vector (starting from initial sighting) at 11h30m

c. what will be its distance and bearingfrom the cutter

Homework Equations

law of sines , law of cosines

The Attempt at a Solution

a. using law of cosines I got part a 6.07km and 78.3 degrees WofS
part b I know it is still 78.3 but I am not sure how to find the displacement vector or how to approach part c

 

[anathema]

For part b, you can use the displacement vector formula: displacement = final position - initial position. In this case, the final position is 4.2km on a bearing of 33 degrees East of North, and the initial position is 9.5km on a bearing of 60 degrees East of North. So the displacement vector would be (4.2cos33 - 9.5cos60, 4.2sin33 - 9.5sin60) = (-3.07, 0.86) km.

For part c, you can use the distance formula d = √(x^2 + y^2) where x and y are the components of the displacement vector. In this case, the distance would be √((-3.07)^2 + (0.86)^2) = 3.2 km. To find the bearing, you can use the inverse tangent function tan^-1(y/x) where x and y are the components of the displacement vector. So the bearing would be tan^-1(0.86/-3.07) = -16.2 degrees East of North.

 

[Moetasim]

As a scientist, it is important to note that the displacement vector of the ship can be represented as a vector quantity, with both magnitude and direction. In this scenario, we can use the information provided to calculate the displacement vector of the ship at 11h10m from its position at 10h30m.

To find the displacement vector, we can use the law of cosines, as you have done, to calculate the magnitude of the vector. However, to find the direction of the vector, we can use the law of sines. This will give us the angle between the initial position of the ship and its final position, which is the direction of the displacement vector.

For part b, we can assume that the ship continues on the same course at the same speed, which means that the magnitude of the displacement vector will remain the same. However, the direction will change as the ship moves from its initial position to its final position. To find the new direction, we can use the law of sines again, using the known magnitude and the new angle.

For part c, we can use the displacement vector calculated in part b to find the distance and bearing from the cutter. This can be done by using the law of cosines to find the distance between the cutter and the ship, and the law of sines to find the bearing from the cutter.

Overall, it is important to use the appropriate equations and pay attention to the units of the given values to accurately calculate the displacement vector and other related quantities.