A particle's position as a function of time is described as `y (t) = 2t^(2) + 3t + 4`. What is the average velocity of the particle from `t = 0` to `t = 3` sec ?

Options

A. 3 m/sec
B. 6 m/sec
C. 9 m/sec
D. 12 m/sec

Explanation

To find the average velocity of the particle from \( t = 0 \) to \( t = 3 \) seconds, we can follow these steps:

### Step 1: Identify the position function
The position of the particle as a function of time is given by:
\[ y(t) = 2t^2 + 3t + 4 \]

### Step 2: Calculate the position at \( t = 0 \) seconds
To find the position at \( t = 0 \) seconds, substitute \( t = 0 \) into the position function:
\[ y(0) = 2(0)^2 + 3(0) + 4 = 4 \text{ meters} \]
Thus, \( y_1 = 4 \) meters.

### Step 3: Calculate the position at \( t = 3 \) seconds
Next, we calculate the position at \( t = 3 \) seconds by substituting \( t = 3 \) into the position function:
\[ y(3) = 2(3)^2 + 3(3) + 4 \]
Calculating this step-by-step:
- \( 3^2 = 9 \)
- \( 2 \times 9 = 18 \)
- \( 3 \times 3 = 9 \)
- Now, add these results: \( 18 + 9 + 4 = 31 \text{ meters} \)

Thus, \( y_2 = 31 \) meters.

### Step 4: Calculate the average velocity
The formula for average velocity \( v_{\text{avg}} \) is given by:
\[ v_{\text{avg}} = \frac{y_2 - y_1}{t_2 - t_1} \]
Here, \( t_1 = 0 \) seconds and \( t_2 = 3 \) seconds. Substituting the values we found:
\[ v_{\text{avg}} = \frac{31 \text{ meters} - 4 \text{ meters}}{3 \text{ seconds} - 0 \text{ seconds}} = \frac{27 \text{ meters}}{3 \text{ seconds}} = 9 \text{ meters/second} \]

### Final Answer
The average velocity of the particle from \( t = 0 \) to \( t = 3 \) seconds is:
\[ \boxed{9 \text{ m/s}} \]

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