What displacement at an angle `60^(@)` to the x-axis has an x-component of 5`m`? `hat(i)` and `hat(j)` are unit vector in x and y directions, respectively.

Options

A. `5hat(i)`
B. `5hat(i)+5hat(j)`
C. `5hat(i)+5sqrt(3)hat(j)`
D. All of the above

Explanation

To find the displacement vector at an angle of \(60^\circ\) to the x-axis with an x-component of \(5 \, m\), we can follow these steps:

### Step 1: Understand the Components of the Displacement
The displacement vector \( \vec{D} \) can be expressed in terms of its components along the x and y axes:
\[ \vec{D} = D_x \hat{i} + D_y \hat{j} \]
where \(D_x\) is the x-component and \(D_y\) is the y-component.

### Step 2: Use the Given Information
We know that the angle \( \theta \) is \(60^\circ\) and the x-component \(D_x\) is \(5 \, m\). The relationship between the components and the angle is given by:
\[ D_x = D \cos(60^\circ) \]
where \(D\) is the magnitude of the displacement.

### Step 3: Calculate the Magnitude of the Displacement
Substituting the known values into the equation:
\[ 5 = D \cos(60^\circ) \]
Since \( \cos(60^\circ) = \frac{1}{2} \):
\[ 5 = D \cdot \frac{1}{2} \]
To find \(D\), multiply both sides by \(2\):
\[ D = 10 \, m \]

### Step 4: Find the y-component of the Displacement
Now that we have the magnitude of the displacement, we can find the y-component using:
\[ D_y = D \sin(60^\circ) \]
Substituting the value of \(D\):
\[ D_y = 10 \sin(60^\circ) \]
Since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
\[ D_y = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \, m \]

### Step 5: Write the Displacement Vector
Now we can write the displacement vector:
\[ \vec{D} = D_x \hat{i} + D_y \hat{j} = 5 \hat{i} + 5\sqrt{3} \hat{j} \]

### Final Answer
Thus, the displacement vector is:
\[ \vec{D} = 5 \hat{i} + 5\sqrt{3} \hat{j} \, m \]

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