Three forces `vec(F)_(1)=(hat(i)+4hat(j))N,vec(F)_(2)=(2hat(i)-4hat(k))N` and `vec(F)_(3)=(hat(k)-4hat(i)-2hat(j))N` are applied on an object of mass 1 kg at rest at origin. The position of the object at t = 2s will be -

Options

A. `(–2m, –6m) `
B. `(–4m, 8m) `
C. `(3m, 6m) `
D. `(2m, – 3m)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the Net Force
We are given three forces acting on the object:
1. \(\vec{F}_1 = \hat{i} + 4\hat{j}\) N
2. \(\vec{F}_2 = 2\hat{i} - 4\hat{k}\) N
3. \(\vec{F}_3 = \hat{k} - 4\hat{i} - 2\hat{j}\) N

To find the net force \(\vec{F}_{net}\), we will add these vectors together:

\[ \vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 \]

Calculating the components:
- **i-component**: \(1 + 2 - 4 = -1\)
- **j-component**: \(4 - 2 = 2\)
- **k-component**: \(-4 + 1 = -3\)

Thus, the net force is:
\[ \vec{F}_{net} = -\hat{i} + 2\hat{j} - 3\hat{k} \text{ N} \]

### Step 2: Calculate the Acceleration
Using Newton's second law, \(\vec{F} = m\vec{a}\), where \(m = 1 \text{ kg}\):

\[ \vec{a} = \frac{\vec{F}_{net}}{m} = \vec{F}_{net} = -\hat{i} + 2\hat{j} - 3\hat{k} \text{ m/s}^2 \]

### Step 3: Calculate the Displacement after 2 seconds
Since the object is initially at rest, we can use the equation of motion:
\[ \vec{s} = \vec{u}t + \frac{1}{2}\vec{a}t^2 \]
where \(\vec{u} = 0\) (initial velocity), \(t = 2 \text{ s}\), and \(\vec{a} = -\hat{i} + 2\hat{j} - 3\hat{k}\).

Substituting the values:
\[ \vec{s} = 0 + \frac{1}{2}(-\hat{i} + 2\hat{j} - 3\hat{k})(2^2) \]
\[ \vec{s} = \frac{1}{2}(-\hat{i} + 2\hat{j} - 3\hat{k})(4) \]
\[ \vec{s} = 2(-\hat{i} + 2\hat{j} - 3\hat{k}) = -2\hat{i} + 4\hat{j} - 6\hat{k} \]

### Step 4: Determine the Position Vector
Since the object starts at the origin \((0, 0, 0)\), the position vector \(\vec{r}\) at \(t = 2 \text{ s}\) is:
\[ \vec{r} = \vec{s} = -2\hat{i} + 4\hat{j} - 6\hat{k} \]

### Final Answer
The position of the object at \(t = 2\) seconds is:
\[ \vec{r} = -2\hat{i} + 4\hat{j} - 6\hat{k} \]

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