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.