The velocity of an object is given by ` vecv = ( 6 t^(3) hati + t^(2) hatj) m//s` . Find the acceleration at t = 2s.

Answer

To find the acceleration of the object at \( t = 2 \) seconds, we will follow these steps:

### Step 1: Write down the given velocity vector.
The velocity vector is given by:
\[ \vec{v} = (6t^3 \hat{i} + t^2 \hat{j}) \, \text{m/s} \]

### Step 2: Differentiate the velocity vector to find the acceleration vector.
Acceleration is defined as the time derivative of the velocity vector:
\[ \vec{a} = \frac{d\vec{v}}{dt} \]

Now, we differentiate the components of the velocity vector:
- For the \( \hat{i} \) component:
\[ \frac{d}{dt}(6t^3) = 6 \cdot 3t^2 = 18t^2 \]
- For the \( \hat{j} \) component:
\[ \frac{d}{dt}(t^2) = 2t \]

Thus, the acceleration vector becomes:
\[ \vec{a} = (18t^2 \hat{i} + 2t \hat{j}) \, \text{m/s}^2 \]

### Step 3: Substitute \( t = 2 \) seconds into the acceleration vector.
Now we will find the acceleration at \( t = 2 \) seconds:
\[ \vec{a} = (18(2^2) \hat{i} + 2(2) \hat{j}) \]
Calculating each component:
- For the \( \hat{i} \) component:
\[ 18(2^2) = 18 \cdot 4 = 72 \]
- For the \( \hat{j} \) component:
\[ 2(2) = 4 \]

Thus, the acceleration vector at \( t = 2 \) seconds is:
\[ \vec{a} = (72 \hat{i} + 4 \hat{j}) \, \text{m/s}^2 \]

### Step 4: Write the final answer.
The acceleration at \( t = 2 \) seconds is:
\[ \vec{a} = 72 \hat{i} + 4 \hat{j} \, \text{m/s}^2 \]

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