A particle of mass 3 kg moves under a froce of `[4 hati + 8 hatj+ 10 hatk]` newton . Calcutale the acceleration (as vector) to which the particle is subjected. If the particle starts from rest and was at the origin initially. What are its new co - ordinates after 3 s?

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the acceleration
We know that the force acting on the particle is given by:
\[ \vec{F} = 4 \hat{i} + 8 \hat{j} + 10 \hat{k} \text{ N} \]
The mass of the particle is:
\[ m = 3 \text{ kg} \]
According to Newton's second law, the acceleration \(\vec{a}\) can be calculated using the formula:
\[ \vec{F} = m \vec{a} \implies \vec{a} = \frac{\vec{F}}{m} \]
Substituting the values:
\[ \vec{a} = \frac{4 \hat{i} + 8 \hat{j} + 10 \hat{k}}{3} \]
Calculating this gives:
\[ \vec{a} = \frac{4}{3} \hat{i} + \frac{8}{3} \hat{j} + \frac{10}{3} \hat{k} \text{ m/s}^2 \]

### Step 2: Determine the new coordinates after 3 seconds
Since the particle starts from rest, the initial velocity \(\vec{u}\) is:
\[ \vec{u} = 0 \text{ m/s} \]
We will use the equation of motion to find the displacement \(\vec{s}\):
\[ \vec{s} = \vec{u} t + \frac{1}{2} \vec{a} t^2 \]
Substituting the values:
\[ \vec{s} = 0 \cdot 3 + \frac{1}{2} \left(\frac{4}{3} \hat{i} + \frac{8}{3} \hat{j} + \frac{10}{3} \hat{k}\right) (3^2) \]
Calculating \(3^2\):
\[ 3^2 = 9 \]
Now substituting this back into the equation:
\[ \vec{s} = \frac{1}{2} \left(\frac{4}{3} \hat{i} + \frac{8}{3} \hat{j} + \frac{10}{3} \hat{k}\right) \cdot 9 \]
This simplifies to:
\[ \vec{s} = \frac{9}{2} \left(\frac{4}{3} \hat{i} + \frac{8}{3} \hat{j} + \frac{10}{3} \hat{k}\right) \]
Calculating each component:
\[ \vec{s} = \left( \frac{9 \cdot 4}{6} \hat{i} + \frac{9 \cdot 8}{6} \hat{j} + \frac{9 \cdot 10}{6} \hat{k} \right) \]
This gives:
\[ \vec{s} = 6 \hat{i} + 12 \hat{j} + 15 \hat{k} \text{ m} \]

### Final Answer
The new coordinates of the particle after 3 seconds are:
\[ \vec{s} = 6 \hat{i} + 12 \hat{j} + 15 \hat{k} \text{ m} \]

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