A current in a wire is given by the equation, . `I=2t^(2)-3t+1`, the charge through cross section of : wire in time interval t=3s to t=5s is

Options

A. 32.33C
B. 43.34C
C. 45.5C
D. 42C

Explanation

To find the charge that flows through the cross-section of a wire in the time interval from \( t = 3 \) seconds to \( t = 5 \) seconds, given the current \( I(t) = 2t^2 - 3t + 1 \), we can follow these steps:

### Step 1: Understand the relationship between current and charge
The current \( I \) is defined as the rate of flow of charge \( Q \) with respect to time \( t \):
\[ I = \frac{dQ}{dt} \]
This means that to find the total charge \( Q \) that flows through the wire over a time interval, we need to integrate the current over that interval.

### Step 2: Set up the integral for charge
To find the charge \( Q \) that flows from \( t = 3 \) seconds to \( t = 5 \) seconds, we can express this as:
\[ Q = \int_{t=3}^{t=5} I(t) \, dt \]
Substituting the expression for current:
\[ Q = \int_{3}^{5} (2t^2 - 3t + 1) \, dt \]

### Step 3: Calculate the integral
Now we will compute the integral:
\[ Q = \int (2t^2 - 3t + 1) \, dt \]
The antiderivative of \( 2t^2 \) is \( \frac{2}{3}t^3 \), the antiderivative of \( -3t \) is \( -\frac{3}{2}t^2 \), and the antiderivative of \( 1 \) is \( t \). Thus:
\[ Q = \left[ \frac{2}{3}t^3 - \frac{3}{2}t^2 + t \right]_{3}^{5} \]

### Step 4: Evaluate the definite integral
Now we evaluate this expression at the limits \( t = 5 \) and \( t = 3 \):
1. Calculate at \( t = 5 \):
\[ Q(5) = \frac{2}{3}(5^3) - \frac{3}{2}(5^2) + 5 \]
\[ = \frac{2}{3}(125) - \frac{3}{2}(25) + 5 \]
\[ = \frac{250}{3} - \frac{75}{2} + 5 \]
Convert \( 5 \) to a fraction:
\[ = \frac{250}{3} - \frac{75}{2} + \frac{15}{3} \]
Find a common denominator (6):
\[ = \frac{500}{6} - \frac{225}{6} + \frac{30}{6} = \frac{305}{6} \]

2. Calculate at \( t = 3 \):
\[ Q(3) = \frac{2}{3}(3^3) - \frac{3}{2}(3^2) + 3 \]
\[ = \frac{2}{3}(27) - \frac{3}{2}(9) + 3 \]
\[ = \frac{54}{3} - \frac{27}{2} + 3 \]
Convert \( 3 \) to a fraction:
\[ = \frac{54}{3} - \frac{27}{2} + \frac{6}{2} = \frac{54}{3} - \frac{21}{2} \]
Find a common denominator (6):
\[ = \frac{108}{6} - \frac{63}{6} = \frac{45}{6} \]

### Step 5: Subtract the two results
Now we find the total charge \( Q \):
\[ Q = Q(5) - Q(3) = \frac{305}{6} - \frac{45}{6} = \frac{260}{6} = \frac{130}{3} \approx 43.33 \text{ coulombs} \]

### Final Answer:
The charge through the cross-section of the wire from \( t = 3 \) seconds to \( t = 5 \) seconds is approximately \( 43.33 \) coulombs.