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The problem asks to calculate the total amount of charge transferred through a conductor over a period of 17 seconds. Here's how the solution is derived step by step:

### Step 1: Understanding the current behavior
- **From 0 to 3 seconds**: The current \( i(t) \) increases linearly from 0 A to 2 A.
- **From 3 to 9 seconds**: The current stays constant at 2 A.
- **From 9 to 17 seconds**: The current decreases linearly from 2 A to 1.5 A.

### Step 2: Applying the relationship between charge and current
The total charge \( q(t) \) is given by the integral of the current \( i(t) \) over time:
\[
q(t) = \int_0^T i(t) dt
\]
where \( i(t) \) is the current as a function of time and \( T \) is the total time (17 seconds in this case).

To compute \( q(t) \), we split the integral into three intervals corresponding to the three different phases of current behavior.

### Step 3: Define the current function \( i(t) \)
The current \( i(t) \) over time is defined piecewise as:
\[
i(t) =
\begin{cases} 
\frac{2t}{3}, & 0 \leq t < 3 \\
2, & 3 \leq t < 9 \\
\left( -\frac{t}{16} + \frac{41}{16} \right), & 9 \leq t \leq 17 
\end{cases}
\]
This function corresponds to:
- A linear increase from 0 to 2 A between 0 and 3 seconds.
- A constant 2 A current between 3 and 9 seconds.
- A linear decrease from 2 A to 1.5 A between 9 and 17 seconds.

### Step 4: Integrating for charge in each interval
1. **From 0 to 3 seconds**:
   \[
   q_1 = \int_0^3 \frac{2t}{3} dt = \left[\frac{t^2}{3}\right]_0^3 = 3 \, C
   \]
   This represents the charge transferred as the current increases from 0 to 2 A.

2. **From 3 to 9 seconds**:
   \[
   q_2 = \int_3^9 2 \, dt = 2 \times (9 - 3) = 12 \, C
   \]
   This represents the charge transferred during the time the current remains constant at 2 A.

3. **From 9 to 17 seconds**:
   \[
   q_3 = \int_9^{17} \left( -\frac{t}{16} + \frac{41}{16} \right) dt = \left[ -\frac{t^2}{32} + \frac{41t}{16} \right]_9^{17} = 14 \, C
   \]
   This represents the charge transferred as the current decreases linearly from 2 A to 1.5 A.

### Step 5: Summing the charge
The total charge transferred is the sum of the charges in all three intervals:
\[
q(t) = q_1 + q_2 + q_3 = 3 \, C + 12 \, C + 14 \, C = 29 \, C
\]

### Conclusion
The total amount of charge transferred over 17 seconds is \( 29 \, C \).

---

Would you like more details on any of these steps or have any specific questions?

### Related Questions
1. How do you derive the piecewise current function \( i(t) \)?
2. What is the physical interpretation of the integral of current over time?
3. Can you explain why the current decreases linearly from 9 to 17 seconds?
4. How does constant current affect charge transfer compared to a varying current?
5. What if the current decreased to 0 instead of 1.5 A after 9 seconds?

### Tip:
Remember that charge is the cumulative effect of current over time, so integrating the current function gives the total charge transferred over the interval.

The current in a conductor changes uniformly from 0 to 2 amps in 3 seconds, remains steady at 2 amps for 6 seconds, and then drops uniformly to 1.5 amps in 8 seconds. Calculate the total amount of charge transferred in the time of 17 seconds.

Learn how to calculate the total charge transferred in a conductor over 17 seconds as the current changes from 0 to 2 amps, stays constant, and then decreases. This solution includes the use of integration and piecewise functions.

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