A vector B, with a magnitude of 4.9 m. is added to a vector A. which lies along an x axis. The sum of these two vectors is a third vector that lies along the y axis and has a magnitude that is twice the magnitude of A. What is the magnitude of A?

Let's break this problem down step by step.

### Step 1: Define the Vectors
We know:
- Vector \( \mathbf{A} \) lies along the x-axis.
- Vector \( \mathbf{B} \) has a magnitude of 4.9 m but is not aligned along either axis (its direction is unknown for now).
- The resultant vector \( \mathbf{R} \) lies along the y-axis and has a magnitude that is twice the magnitude of \( \mathbf{A} \).

### Step 2: Use the Given Information
Let the magnitude of vector \( \mathbf{A} \) be \( A \).

So, \( |\mathbf{R}| = 2A \), since the resultant vector has a magnitude that is twice that of \( \mathbf{A} \).

Now, since \( \mathbf{R} \) lies entirely along the y-axis, the x-component of the resultant must be 0. Therefore, the x-components of \( \mathbf{A} \) and \( \mathbf{B} \) must cancel each other out.

### Step 3: Decompose the Vectors
- \( \mathbf{A} \) only has an x-component: \( \mathbf{A} = (A, 0) \).
- \( \mathbf{B} \) has both an x-component and a y-component. Let the components of \( \mathbf{B} \) be \( (B_x, B_y) \).

The sum of the vectors is:
\[
\mathbf{R} = \mathbf{A} + \mathbf{B}
\]

So, in component form:
\[
(0, 2A) = (A + B_x, B_y)
\]

This gives us two equations:
1. \( A + B_x = 0 \) (x-component must cancel out)
2. \( B_y = 2A \) (y-component results in the magnitude of \( \mathbf{R} \))

### Step 4: Solve for the Magnitude of \( A \)
From the first equation, \( B_x = -A \).

Now, use the fact that the magnitude of \( \mathbf{B} \) is 4.9 m:
\[
|\mathbf{B}| = \sqrt{B_x^2 + B_y^2}
\]
Substitute \( B_x = -A \) and \( B_y = 2A \):
\[
4.9 = \sqrt{(-A)^2 + (2A)^2}
\]
\[
4.9 = \sqrt{A^2 + 4A^2}
\]
\[
4.9 = \sqrt{5A^2}
\]
\[
4.9 = A\sqrt{5}
\]
Solve for \( A \):
\[
A = \frac{4.9}{\sqrt{5}} = \frac{4.9}{2.236} \approx 2.19 \, \text{m}
\]

Thus, the magnitude of \( A \) is approximately **2.19 m**.

### Final Answer:
The magnitude of vector \( A \) is approximately 2.19 m.

Do you have any questions or need further clarification? Here are five related questions for further practice:

1. What is the direction of vector \( B \) with respect to the x-axis?
2. How would the problem change if the resultant vector had a different magnitude?
3. How can you determine the angle between vectors \( A \) and \( B \)?
4. What happens if vector \( A \) had a negative x-component?
5. How would the result differ if the magnitude of vector \( B \) were doubled?

**Tip**: Always break down vector problems into their components to simplify the calculations.

A vector B, with a magnitude of 4.9 m, is added to a vector A, which lies along an x axis. The sum of these two vectors is a third vector that lies along the y axis and has a magnitude that is twice the magnitude of A. What is the magnitude of A?

This solution breaks down how to calculate the magnitude of vector A when a vector B with a magnitude of 4.9 m is added to it, resulting in a vector that lies along the y-axis with twice the magnitude of A. The step-by-step guide covers vector addition, component decomposition, and the use of the Pythagorean theorem, suitable for Grades 11-12.

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