Magnitude and direction of ship relative to ground

[physicsquest]

Homework Statement

At the entrance of Ambrose Channel at New York harbor, the tidal current at one time of the day has a velocity of 4.2 km/h in a direction 20 degrees south of east. Consider a ship in this current; suppose that the ship has a speed of 16 km/h relative to the water. If the helmsman keeps the bow of the ship aimed due north, what will be the actual velocity (magnitude and direction) of the ship relative to the ground?

Homework Equations

vx= vtx-vsx
vy= vty-vsy
tan$$\phi$$= [tex]\sqrt{(vx^2 + vy^2)}
tan\phi=vy/vx

The Attempt at a Solution

 

[tiny-tim]

At the entrance of Ambrose Channel at New York harbor, the tidal current at one time of the day has a velocity of 4.2 km/h in a direction 20 degrees south of east. Consider a ship in this current; suppose that the ship has a speed of 16 km/h relative to the water. If the helmsman keeps the bow of the ship aimed due north, what will be the actual velocity (magnitude and direction) of the ship relative to the ground?

Hi physicsquest!

Relative velocities are vectors, so draw a vector triangle …

what is the length and direction of the third side? ​

 

[tomcue]

To calculate the actual velocity of the ship relative to the ground, we need to consider both the velocity of the ship relative to the water and the velocity of the water relative to the ground. We can use vector addition to find the resultant velocity.

First, we need to convert all velocities to the same units, which in this case is km/h. The velocity of the water relative to the ground is given as 4.2 km/h in a direction 20 degrees south of east. This means that the x-component of the water's velocity is 4.2*cos(20)= 3.98 km/h and the y-component is 4.2*sin(20)= 1.43 km/h.

Next, we need to find the x and y components of the ship's velocity relative to the water. The ship's speed relative to the water is given as 16 km/h, and since the ship is aimed due north, the x-component of the ship's velocity will be 0 km/h and the y-component will be 16 km/h.

Now, we can use vector addition to find the resultant velocity of the ship relative to the ground. The x-component of the resultant velocity will be the sum of the x-components of the ship and water velocities, which is 3.98+0= 3.98 km/h. The y-component will be the sum of the y-components, which is 1.43+16= 17.43 km/h.

To find the magnitude and direction of the resultant velocity, we can use the equations given in the homework section. The magnitude of the resultant velocity is given by:
[tex]\sqrt{(3.98^2 + 17.43^2)}= 17.89 km/h

The direction of the resultant velocity can be found using the equation tan\phi=vy/vx, where phi is the angle between the resultant velocity and the x-axis. Therefore, tan\phi= 17.43/3.98= 4.38, and solving for phi, we get phi= 76.77 degrees.

Therefore, the actual velocity of the ship relative to the ground is 17.89 km/h in a direction 76.77 degrees north of east. This means that the ship will be moving at a faster speed and in a slightly different direction compared to its initial velocity relative to the water, due to the influence of the tidal current.