A ship leaves port and sails northwest for 2 hours, and then sails N35degreesE for three hours. If it does not change speed, what direction must the ship sail to return directly to port? How long will this return trip take?
To determine the direction the ship must sail to return directly to port, we can analyze the problem using vector addition.
First, let's represent the ship's movement as vectors on a coordinate plane. We can consider the initial northwest movement as vector A and the subsequent N35°E movement as vector B.
Vector A can be divided into two components: a northward component and a westward component. The westward component cancels out the eastward component of vector B, leaving only the northward component of vector A and the northward component of vector B.
Next, we need to find the resultant vector of the northward components. To do this, we can add the magnitudes of the northward components and their directions as angles from the north.
The northward component of vector A can be calculated using trigonometry. Since northwest is halfway between north and west, the northward component of vector A is given by: A*cos(45°).
The northward component of vector B is given by: B*sin(35°).
Now, let's find the resultant vector's magnitude and direction. The magnitude can be obtained by summing the magnitudes of the northward components:
Resultant magnitude = A*cos(45°) + B*sin(35°).
To find the direction, we can use the tangent function:
Resultant direction = arctan((B*sin(35°))/(A*cos(45°))).
This resultant direction will give us the angle with respect to the north direction, which can be converted into a compass heading.
To determine the time it takes to return to port, we would need the speed of the ship. Since it is mentioned in the problem that the ship does not change speed, we can assume it remains constant. Therefore, the return trip will take the same amount of time as the initial trip.
By calculating the resultant direction and having the speed of the ship, you can determine the exact compass heading and the time it will take for the return trip.
2h[135o] + 3h[55o]
X = 2*Cos135 + 3*Cos55 = 0.307 h.
Y = 2*sin135 + 3*sin55 = 3.87 h.
Tan A = Y/X = 3.87/0.307 = 12.61
A = 85.5o = The direction.
T = Y/sin85.5 = 3.88 h.