An ocean liner leaves New York City and travels 25.1 degrees north of east for 110km. How far east and how far north has it gone? In other words, what are the magnitudes of the components of the ships displacement vector in the directions, a) due east and b) due north
Trig, not physics
Note they gave you math class angle measured north of east (counterclockwise from x axis), not compass angle as used by navigator or physicist.
east 110 cos 25.1
north 110 sin 25.1
Well, well, well, seems like the ocean liner has gone on a little adventure! Let's not keep it waiting for the answers, shall we?
a) Due east, we can use some trigonometry to find the magnitude of the eastward component. Considering that it has traveled at a 25.1-degree angle north of east, we can decompose the displacement vector into its horizontal and vertical components using some cosine magic.
The magnitude of the eastward component (cosine of the angle times the total displacement) would be:
cos(25.1) * 110 km.
b) Due north, we can again use some trigonometry magic to find the magnitude of the northward component. This time we'll use the sine function.
The magnitude of the northward component (sine of the angle times the total displacement) would be:
sin(25.1) * 110 km.
I hope this helps, and don't forget to tell the ocean liner I said "bon voyage"!
To find the magnitudes of the components of the ship's displacement vector, we can use trigonometry.
a) Due East:
The ship travels 25.1 degrees north of east, which means it is heading slightly upwards. To find the east component, we need to find the cosine of this angle and multiply it by the total distance traveled.
East Component = Distance Traveled * cos(angle)
East Component = 110 km * cos(25.1 degrees)
East Component ≈ 99.98 km
Therefore, the magnitude of the component of the ship's displacement in the due east direction is approximately 99.98 km.
b) Due North:
The ship is not directly heading north, but we can still find the north component using trigonometry. To find the north component, we need to find the sine of the angle and multiply it by the total distance traveled.
North Component = Distance Traveled * sin(angle)
North Component = 110 km * sin(25.1 degrees)
North Component ≈ 46.81 km
Therefore, the magnitude of the component of the ship's displacement in the due north direction is approximately 46.81 km.
To determine the magnitudes of the components of the ship's displacement vector in the east and north directions, we can use trigonometry.
Let's break down the information given:
- The ocean liner travels 25.1 degrees north of east.
- The total distance traveled is 110 km.
We can use the concept of vectors to represent the ship's displacement. In this case, the displacement vector is the straight-line distance and direction from the starting point (New York City) to the final point.
To find the magnitude of the component of the displacement vector in the east direction (a), we need to calculate the projection of the displacement vector onto the eastward axis. The formula for this is:
Magnitude east = Total distance * cosine of the angle
Magnitude east = 110 km * cos(25.1°)
Calculating this will give you the distance traveled east by the ocean liner.
Similarly, to find the magnitude of the component in the north direction (b), we need to calculate the projection of the displacement vector onto the northward axis. The formula for this is:
Magnitude north = Total distance * sine of the angle
Magnitude north = 110 km * sin(25.1°)
By calculating this, you will find the distance traveled north by the ocean liner.
Remember to use a scientific or graphing calculator for the trigonometric calculations.