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 start with the given velocity vector:

\[ \vec{v} = (6t^3 \hat{i} + t^2 \hat{j}) \, \text{m/s} \]

### Step 1: Differentiate the velocity vector to find acceleration
The acceleration vector \( \vec{a} \) is given by the derivative of the velocity vector with respect to time:

\[ \vec{a} = \frac{d\vec{v}}{dt} \]

### Step 2: Differentiate each component of the velocity vector
We differentiate the components of the velocity vector:

1. For the \( \hat{i} \) component:
\[ \frac{d}{dt}(6t^3) = 18t^2 \]

2. For the \( \hat{j} \) component:
\[ \frac{d}{dt}(t^2) = 2t \]

Combining these results, we have:

\[ \vec{a} = (18t^2 \hat{i} + 2t \hat{j}) \, \text{m/s}^2 \]

### Step 3: Substitute \( t = 2 \) seconds into the acceleration vector
Now we substitute \( t = 2 \) seconds into the acceleration vector:

1. For the \( \hat{i} \) component:
\[ 18(2^2) = 18 \times 4 = 72 \]

2. For the \( \hat{j} \) component:
\[ 2(2) = 4 \]

Thus, the acceleration vector at \( t = 2 \) seconds is:

\[ \vec{a} = (72 \hat{i} + 4 \hat{j}) \, \text{m/s}^2 \]

### 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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