A particle moves in a straight line. Its position ( in m) as function of time is given by
`x = (at^2 + b)`
What is the average velocity in time interval ` t = 3s to t = 5s in ms^(-1)`. (where a and b are constants and a `= 1ms^(-2), b = 1m`).

Explanation

To find the average velocity of the particle in the time interval from \( t = 3 \, \text{s} \) to \( t = 5 \, \text{s} \), we can follow these steps:

### Step 1: Write down the position function
The position of the particle as a function of time is given by:
\[ x(t) = at^2 + b \]
where \( a = 1 \, \text{m/s}^2 \) and \( b = 1 \, \text{m} \). Substituting the values of \( a \) and \( b \):
\[ x(t) = 1t^2 + 1 = t^2 + 1 \]

### Step 2: Calculate the position at \( t = 3 \, \text{s} \)
Now, we need to find the position of the particle at \( t = 3 \, \text{s} \):
\[ x(3) = (3)^2 + 1 = 9 + 1 = 10 \, \text{m} \]

### Step 3: Calculate the position at \( t = 5 \, \text{s} \)
Next, we find the position of the particle at \( t = 5 \, \text{s} \):
\[ x(5) = (5)^2 + 1 = 25 + 1 = 26 \, \text{m} \]

### Step 4: Calculate the change in position
Now we calculate the change in position (\( \Delta x \)) between \( t = 3 \, \text{s} \) and \( t = 5 \, \text{s} \):
\[ \Delta x = x(5) - x(3) = 26 \, \text{m} - 10 \, \text{m} = 16 \, \text{m} \]

### Step 5: Calculate the change in time
The change in time (\( \Delta t \)) is:
\[ \Delta t = 5 \, \text{s} - 3 \, \text{s} = 2 \, \text{s} \]

### Step 6: Calculate the average velocity
The average velocity (\( V_{\text{avg}} \)) is given by the formula:
\[ V_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{16 \, \text{m}}{2 \, \text{s}} = 8 \, \text{m/s} \]

### Step 7: Convert to m/s
Since the answer is required in meters per second, we can write:
\[ V_{\text{avg}} = 8 \, \text{m/s} \]

### Final Answer
The average velocity of the particle in the time interval from \( t = 3 \, \text{s} \) to \( t = 5 \, \text{s} \) is \( 8 \, \text{m/s} \).

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