Vectors
A river has a current with a velocity of 6.45 m/s that flows due south.  A person paddling in a kayak wishes to cross this river with a net velocity of 4.35 m/s due east.
a) Calculate the angle that the paddler must point the kayak.  
Assuming the positive y-axis points north and the positive x-axis points east, give your answer using the standard convention for angle measurements.  
Answer: Answer for part 1o
b) Calculate the speed at which the kayaker must paddle.
Answer: Answer for part 2 m/s


### Part (a): Calculating the Angle

We have the following vectors:
- The river’s current velocity: \( \mathbf{v}_{\text{current}} = 6.45 \, \text{m/s} \) (southward, which corresponds to the negative y-axis).
- The desired net velocity: \( \mathbf{v}_{\text{net}} = 4.35 \, \text{m/s} \) (due east, along the positive x-axis).

Let the velocity of the paddler relative to the water be \( \mathbf{v}_{\text{paddler}} \), and let \( \theta \) be the angle the paddler must point the kayak relative to the east (x-axis). We need to calculate this angle.

The velocity components of the paddler relative to the river can be broken down into:
- \( v_{\text{paddler}} \cos(\theta) \) in the eastward (x) direction.
- \( v_{\text{paddler}} \sin(\theta) \) in the northward (y) direction.

The net eastward velocity must be \( 4.35 \, \text{m/s} \), and the southward current is \( 6.45 \, \text{m/s} \), which will require a northward paddling velocity to counteract it. Therefore, we have two equations:

\[
v_{\text{paddler}} \cos(\theta) = 4.35 \, \text{m/s} \quad \text{(1)}
\]
\[
v_{\text{paddler}} \sin(\theta) = 6.45 \, \text{m/s} \quad \text{(2)}
\]

Dividing equation (2) by equation (1) to eliminate \( v_{\text{paddler}} \), we get:
\[
\tan(\theta) = \frac{6.45}{4.35}
\]
\[
\theta = \tan^{-1}\left( \frac{6.45}{4.35} \right)
\]
\[
\theta \approx \tan^{-1}(1.4828) \approx 56.56^\circ
\]

Thus, the kayaker must point the kayak at an angle of **\( 56.56^\circ \)** north of east.

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### Part (b): Calculating the Speed at Which the Kayaker Must Paddle

To find the speed at which the kayaker must paddle, we can use the Pythagorean theorem, combining the two velocity components:

\[
v_{\text{paddler}} = \sqrt{(v_{\text{paddler}} \cos(\theta))^2 + (v_{\text{paddler}} \sin(\theta))^2}
\]

From equations (1) and (2):
\[
v_{\text{paddler}} = \sqrt{4.35^2 + 6.45^2}
\]
\[
v_{\text{paddler}} = \sqrt{18.9225 + 41.6025}
\]
\[
v_{\text{paddler}} \approx \sqrt{60.525} \approx 7.78 \, \text{m/s}
\]

Thus, the kayaker must paddle at a speed of **7.78 m/s**.

---

Would you like more details on these calculations or further explanations?

Here are 5 related questions:
1. How would the answer change if the current flowed north instead of south?
2. What happens to the paddling angle if the river current increases?
3. How do you calculate the net velocity if the kayaker paddles at a different angle?
4. What is the relation between velocity vectors and their components?
5. How can trigonometric functions be applied to resolve vectors?

**Tip**: When working with vectors, always break them into components along the axes to simplify calculations.

A river has a current with a velocity of 6.45 m/s that flows due south. A person paddling in a kayak wishes to cross this river with a net velocity of 4.35 m/s due east. a) Calculate the angle that the paddler must point the kayak. Assuming the positive y-axis points north and the positive x-axis points east, give your answer using the standard convention for angle measurements. b) Calculate the speed at which the kayaker must paddle.

Solve a vector problem where a kayaker paddles across a river with a current flowing south at 6.45 m/s and a desired velocity of 4.35 m/s due east. Learn how to calculate the angle at which the kayaker must point and the required paddling speed using trigonometric formulas and the Pythagorean theorem. Suitable for Grades 10-12.

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