I am having trouble understanding the reasoning behind the solution in this Irodov General Physics problem. The problem is 1.6:

1.6. A ship moves along the equator to the east with velocity vo = 30 km/hour. The southeastern wind blows at an angle Θ = 60° to the equator with velocity v = 15 km/hour. Find the wind velocity v' relative to the ship and the angle Θ' between the equator and the wind direction in the reference frame fixed to the ship.

My calculations:

After drawing the initial diagram I consider that in order for the wind direction to be relative to the boat, I need to subtract the boat vector from the wind vector.

In my diagram I have repositioned the boat vector to be subtracted from the wind vector and using trigonometry I solve as follows:

First the cosine law to determine the magnitude

The magnitude = $\sqrt{30^2 + 15^2 - 2\cdot 15\cdot 30\cdot \cos{60}}$

= $\sqrt{675} \approx 25.98$

Then for the angle I can use sine law

$\frac{sin(\Theta)}{30 km/h} = \frac{sin(60)}{\sqrt{675}}$

$\Theta= \arcsin{30km/h \cdot\frac{sin(60)}{\sqrt{675}}}$

$\Theta = 90°$

The answer I get is therefore that the wind, in reference to the boat (as felt by the people standing on the boat), is approx 26 km/h at an angle of 60° South of West (in relation to the equator).

However the answer given in the solutions is always that the approximate speed of the wind is 40km/h and the angle is approx 19°.

I have looked at the solutions and I think the best explanation is given in the following diagrams obtained from link #3:

I do not understand, however, why he is using cos(180+$\Theta$). I want to understand why my solution is incorrect and yields a different answer.

  What is my mistake and what is the proper intuition for understanding this problem if I have done so incorrectly?

Solutions I have viewed:

1. This Youtube Video

2. This online blog

3. This online blog

I am having trouble understanding the reasoning behind the solution in this Irodov General Physics problem:

problem 1.6.) A ship moves along the equator to the east with velocity $v_0=30\;km/h$. The southeastern wind blows at an angle $\theta=60°$ to the equator with velocity $v=15\;km/h$. Find the wind velocity $v'$ relative to the ship and the angle $\theta'$ between the equator and the wind direction in the reference frame fixed to the ship.

My calculations:

After drawing the initial diagram I consider that in order for the wind direction to be relative to the boat, I need to subtract the boat vector from the wind vector.

In my diagram I have repositioned the boat vector to be subtracted from the wind vector and using trigonometry I solve as follows:

First the cosine law to determine the magnitude

The magnitude = $\sqrt{30^2 + 15^2 - 2\cdot 15\cdot 30\cdot \cos{60}}$

= $\sqrt{675} \approx 25.98$

Then for the angle I can use sine law

$\frac{sin(\theta)}{30\frac {km}{h}} = \frac{sin(60)}{\sqrt{675}}$

$\theta= \arcsin{\frac{30\frac {km}{h}\cdot sin(60)}{\sqrt{675}}}$

$\theta = 90°$

The answer I get is therefore that the wind, in reference to the boat (as felt by the people standing on the boat), is approx $26\:km/h$ at an angle of $60°$ South of West (in relation to the equator).

However the answer given in the solutions is always that the approximate speed of the wind is $40\:km/h$ and the angle is approx $19°$.

I have looked at the solutions and I think the best explanation is given in the following diagrams obtained from link #3:

I do not understand, however, why he is using $\cos{(\pi +\theta)}$. I want to understand why my solution is incorrect and yields a different answer. What is my mistake and what is the proper intuition for understanding this problem if I have done so incorrectly?

Solutions I have viewed:

1. This Youtube Video

2. This online blog

3. This online blog

I am having trouble understanding the reasoning behind the solution in this Irodov General Physics problem. The problem is 1.6:

1.6. A ship moves along the equator to the east with velocity vo = 30 km/hour. The southeastern wind blows at an angle cp = 60° to the equator with velocity v = 15 km/hour. Find the wind velocity v' relative to the ship and the angle p' between the equator and the wind direction in the reference frame fixed to the ship.

My calculations:

After drawing the initial diagram I consider that in order for the wind direction to be relative to the boat, I need to subtract the boat vector from the wind vector.

In my diagram I have repositioned the boat vector to be subtracted from the wind vector and using trigonometry I solve as follows:

First the cosine law to determine the magnitude

The magnitude = $\sqrt{30^2 + 15^2 - 2\cdot 15\cdot 30\cdot \cos{60}}$

= $\sqrt{675} \approx 25.98$

Then for the angle I can use sine law

$\frac{sin(\Theta)}{30 km/h} = \frac{sin(60)}{\sqrt{675}}$

$\Theta= \arcsin{30km/h \cdot\frac{sin(60)}{\sqrt{675}}}$

$\Theta = 90°$

The answer I get is therefore that the wind, in reference to the boat (as felt by the people standing on the boat), is approx 26 km/h at an angle of 60° South of West (in relation to the equator).

However the answer given in the solutions is always that the approximate speed of the wind is 40km/h and the angle is approx 19°.

I have looked at the solutions and I think the best explanation is given in the following diagrams obtained from link #3:

I do not understand, however, why he is using cos(180+$\Theta$). I want to understand why my solution is incorrect and yields a different answer.

What is my mistake and what is the proper intuition for understanding this problem if I have done so incorrectly?

Solutions I have viewed:

1. This Youtube Video

2. This online blog

3. This online blog

I am having trouble understanding the reasoning behind the solution in this Irodov General Physics problem. The problem is 1.6:

1.6. A ship moves along the equator to the east with velocity vo = 30 km/hour. The southeastern wind blows at an angle Θ = 60° to the equator with velocity v = 15 km/hour. Find the wind velocity v' relative to the ship and the angle Θ' between the equator and the wind direction in the reference frame fixed to the ship.

My calculations:

After drawing the initial diagram I consider that in order for the wind direction to be relative to the boat, I need to subtract the boat vector from the wind vector.

In my diagram I have repositioned the boat vector to be subtracted from the wind vector and using trigonometry I solve as follows:

First the cosine law to determine the magnitude

The magnitude = $\sqrt{30^2 + 15^2 - 2\cdot 15\cdot 30\cdot \cos{60}}$

= $\sqrt{675} \approx 25.98$

Then for the angle I can use sine law

$\frac{sin(\Theta)}{30 km/h} = \frac{sin(60)}{\sqrt{675}}$

$\Theta= \arcsin{30km/h \cdot\frac{sin(60)}{\sqrt{675}}}$

$\Theta = 90°$

The answer I get is therefore that the wind, in reference to the boat (as felt by the people standing on the boat), is approx 26 km/h at an angle of 60° South of West (in relation to the equator).

However the answer given in the solutions is always that the approximate speed of the wind is 40km/h and the angle is approx 19°.

I have looked at the solutions and I think the best explanation is given in the following diagrams obtained from link #3:

I do not understand, however, why he is using cos(180+$\Theta$). I want to understand why my solution is incorrect and yields a different answer.

What is my mistake and what is the proper intuition for understanding this problem if I have done so incorrectly?

Solutions I have viewed:

1. This Youtube Video

2. This online blog

3. This online blog

I am having trouble understanding the reasoning behind the solution in this Irodov General Physics problem. The problem is 1.6:

1.6. A ship moves along the equator to the east with velocity vo = 30 km/hour. The southeastern wind blows at an angle cp = 60° to the equator with velocity v = 15 km/hour. Find the wind velocity v' relative to the ship and the angle p' between the equator and the wind direction in the reference frame fixed to the ship.

My calculations:

After drawing the initial diagram I consider that in order for the wind direction to be relative to the boat, I need to subtract the boat vector from the wind vector.

In my diagram I have repositioned the boat vector to be subtracted from the wind vector and using trigonometry I solve as follows:

First the cosine law to determine the magnitude

The magnitude = $\sqrt{30^2 + 15^2 - 2\cdot 15\cdot 30\cdot \cos{60}}$

= $\sqrt{675} \approx 25.98$

Then for the angle I can use sine law

$\frac{sin(\Theta)}{30 km/h} = \frac{sin(60)}{\sqrt{675}}$

$\Theta= \arcsin{30km/h \cdot\frac{sin(60)}{\sqrt{675}}}$

$\Theta = 90°$

The answer I get is therefore that the wind, in reference to the boat (as felt by the people standing on the boat), is approx 26 km/h at an angle of 60° South of West (in relation to the equator).

However the answer given in the solutions is always that the approximate speed of the wind is 40km/h and the angle is approx 19°.

I have looked at the solutions and I think the best explanation is given in the following diagrams obtained from link #3:

I do not understand, however, why he is using cos(180+$\Theta$). I want to understand why my solution is incorrect and yields a different answer.

What is my mistake and what is the proper intuition for understanding this problem if I have done so incorrectly?

Solutions I have viewed:

1. This Youtube Video

2. This online blog

3. This online blog

I am having trouble understanding the reasoning behind the solution in this Irodov General Physics problem. The problem is 1.6:

1.6. A ship moves along the equator to the east with velocity vo = 30 km/hour. The southeastern wind blows at an angle cp = 60° to the equator with velocity v = 15 km/hour. Find the wind velocity v' relative to the ship and the angle p' between the equator and the wind direction in the reference frame fixed to the ship.

My calculations:

After drawing the initial diagram I consider that in order for the wind direction to be relative to the boat, I need to subtract the boat vector from the wind vector.

In my diagram I have repositioned the boat vector to be subtracted from the wind vector and using trigonometry I solve as follows:

First the cosine law to determine the magnitude

The magnitude = $\sqrt{30^2 + 15^2 - 2\cdot 15\cdot 30\cdot \cos{60}}$

= $\sqrt{675} \approx 25.98$

Then for the angle I can use sine law

$\frac{sin(\Theta)}{30 km/h} = \frac{sin(60)}{\sqrt{675}}$

$\Theta= \arcsin{30km/h \cdot\frac{sin(60)}{\sqrt{675}}}$

$\Theta = 90°$

The answer I get is therefore that the wind, in reference to the boat (as felt by the people standing on the boat), is approx 26 km/h at an angle of 60° South of West (in relation to the equator).

However the answer given in the solutions is always that the approximate speed of the wind is 40km/h and the angle is approx 19°.

I have looked at the solutions and I think the best explanation is given in the following diagrams obtained from link #3:

I do not understand, however, why he is using cos(180+$\Theta$). I want to understand why my solution is incorrect and yields a different answer.

What is my mistake and what is the proper intuition for understanding this problem if I have done so incorrectly?

Solutions I have viewed:

1. This Youtube Video

2. This online blog

3. This online blog