Electric current in a wire is time rate of flow of charge. The charge in coulombs that passes through a wire after `t` seconds is given by the function `q(t)=t^(2)-2t^(2)+5t+2`. Determine the average current (in coulmbe per second) during the first two seconds.

Options

A. `4A`
B. `5A`
C. `6A`
D. `12A`

Explanation

To determine the average current during the first two seconds given the charge function \( q(t) = t^3 - 2t^2 + 5t + 2 \), we can follow these steps:

### Step 1: Calculate the charge at \( t = 0 \) seconds
We start by finding the charge \( q(0) \):
\[ q(0) = (0)^3 - 2(0)^2 + 5(0) + 2 = 2 \, \text{C} \]

### Step 2: Calculate the charge at \( t = 2 \) seconds
Next, we calculate the charge \( q(2) \):
\[ q(2) = (2)^3 - 2(2)^2 + 5(2) + 2 \]
Calculating each term:
- \( (2)^3 = 8 \)
- \( -2(2)^2 = -8 \)
- \( 5(2) = 10 \)
- \( +2 = 2 \)

Now, substituting these values:
\[ q(2) = 8 - 8 + 10 + 2 = 12 \, \text{C} \]

### Step 3: Calculate the change in charge (\( \Delta q \))
Now we find the change in charge over the interval from \( t = 0 \) to \( t = 2 \):
\[ \Delta q = q(2) - q(0) = 12 \, \text{C} - 2 \, \text{C} = 10 \, \text{C} \]

### Step 4: Calculate the time interval (\( \Delta t \))
The time interval for this calculation is:
\[ \Delta t = 2 \, \text{s} - 0 \, \text{s} = 2 \, \text{s} \]

### Step 5: Calculate the average current (\( I_{avg} \))
The average current is given by the formula:
\[ I_{avg} = \frac{\Delta q}{\Delta t} \]
Substituting the values we found:
\[ I_{avg} = \frac{10 \, \text{C}}{2 \, \text{s}} = 5 \, \text{A} \]

### Final Answer
The average current during the first two seconds is \( 5 \, \text{A} \).

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