The position of an object is given by ` vecr= ( 9 thati + 4t^(3)hatj) ` m. find its velocity at time t= 1s.

Answer

To find the velocity of an object at a given time when its position is described by the vector equation, we can follow these steps:

### Step 1: Write down the position vector
The position vector of the object is given as:
\[ \vec{r} = 9t \hat{i} + 4t^3 \hat{j} \quad \text{(in meters)} \]

### Step 2: Differentiate the position vector to find the velocity vector
To find the velocity vector \(\vec{v}\), we need to differentiate the position vector \(\vec{r}\) with respect to time \(t\):
\[ \vec{v} = \frac{d\vec{r}}{dt} \]

### Step 3: Differentiate each component
Differentiate the components of \(\vec{r}\):
- The \(x\)-component:
\[ \frac{d}{dt}(9t) = 9 \hat{i} \]
- The \(y\)-component:
\[ \frac{d}{dt}(4t^3) = 12t^2 \hat{j} \]

### Step 4: Combine the differentiated components
Thus, the velocity vector becomes:
\[ \vec{v} = 9 \hat{i} + 12t^2 \hat{j} \]

### Step 5: Substitute \(t = 1\) second into the velocity vector
Now, we substitute \(t = 1\) second into the velocity vector:
\[ \vec{v} = 9 \hat{i} + 12(1)^2 \hat{j} = 9 \hat{i} + 12 \hat{j} \]

### Step 6: Calculate the magnitude of the velocity vector
To find the magnitude of the velocity vector, we use the formula:
\[ |\vec{v}| = \sqrt{(v_x)^2 + (v_y)^2} \]
Substituting the components:
\[ |\vec{v}| = \sqrt{(9)^2 + (12)^2} = \sqrt{81 + 144} = \sqrt{225} = 15 \, \text{m/s} \]

### Final Answer
The velocity of the object at \(t = 1\) second is:
\[ \vec{v} = 9 \hat{i} + 12 \hat{j} \quad \text{with a magnitude of } 15 \, \text{m/s} \]

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