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The problem asks about the velocity of a particle, where the displacement \( s \) as a function of time \( t \) is given by:

\[
s(t) = t^2 - 7t + 15
\]

### (a) Finding the average velocity over each time interval

The formula for **average velocity** over a time interval \([t_1, t_2]\) is:

\[
\text{Average Velocity} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}
\]

Let's calculate the average velocity for each interval:

1. **For interval [3, 4]:**
\[
\text{Average Velocity} = \frac{s(4) - s(3)}{4 - 3}
\]
We will compute \(s(4)\) and \(s(3)\):
\[
s(4) = 4^2 - 7(4) + 15 = 16 - 28 + 15 = 3
\]
\[
s(3) = 3^2 - 7(3) + 15 = 9 - 21 + 15 = 3
\]
Thus,
\[
\text{Average Velocity} = \frac{3 - 3}{4 - 3} = \frac{0}{1} = 0 \, \text{m/s}
\]

2. **For interval [3.5, 4]:**
\[
\text{Average Velocity} = \frac{s(4) - s(3.5)}{4 - 3.5}
\]
We already know \(s(4) = 3\), and now we compute \(s(3.5)\):
\[
s(3.5) = (3.5)^2 - 7(3.5) + 15 = 12.25 - 24.5 + 15 = 2.75
\]
Thus,
\[
\text{Average Velocity} = \frac{3 - 2.75}{4 - 3.5} = \frac{0.25}{0.5} = 0.5 \, \text{m/s}
\]

3. **For interval [4, 5]:**
\[
\text{Average Velocity} = \frac{s(5) - s(4)}{5 - 4}
\]
We already know \(s(4) = 3\), and now we compute \(s(5)\):
\[
s(5) = 5^2 - 7(5) + 15 = 25 - 35 + 15 = 5
\]
Thus,
\[
\text{Average Velocity} = \frac{5 - 3}{5 - 4} = \frac{2}{1} = 2 \, \text{m/s}
\]

4. **For interval [4, 4.5]:**
\[
\text{Average Velocity} = \frac{s(4.5) - s(4)}{4.5 - 4}
\]
We already know \(s(4) = 3\), and now we compute \(s(4.5)\):
\[
s(4.5) = (4.5)^2 - 7(4.5) + 15 = 20.25 - 31.5 + 15 = 3.75
\]
Thus,
\[
\text{Average Velocity} = \frac{3.75 - 3}{4.5 - 4} = \frac{0.75}{0.5} = 1.5 \, \text{m/s}
\]

### (b) Finding the instantaneous velocity at \( t = 4 \)

The **instantaneous velocity** is the derivative of the displacement function \( s(t) \). So, let's differentiate \( s(t) = t^2 - 7t + 15 \):

\[
v(t) = \frac{ds}{dt} = 2t - 7
\]

Now, substitute \( t = 4 \) into the derivative:

\[
v(4) = 2(4) - 7 = 8 - 7 = 1 \, \text{m/s}
\]

### Final Answers:
- (a) Average velocities:
  1. (i) 0 m/s
  2. (ii) 0.5 m/s
  3. (iii) 2 m/s
  4. (iv) 1.5 m/s
- (b) Instantaneous velocity at \( t = 4 \): 1 m/s

Do you need any additional details or clarifications? 

Here are some related questions to extend your understanding:
1. How does the derivative relate to velocity in general?
2. Can you find the instantaneous velocity at other points, like \( t = 5 \)?
3. What is the physical meaning of a zero average velocity over a time interval?
4. How would you find the acceleration of the particle?
5. How would the problem change if the displacement function were cubic?

**Tip:** To find instantaneous velocity, always differentiate the position function with respect to time.

The displacement (in meters) of a particle moving in a straight line is given by s = t^2 − 7t + 15, where t is measured in seconds. (a) Find the average velocity over each time interval: (i) [3,4], (ii) [3.5, 4], (iii) [4, 5], (iv) [4, 4.5]. (b) Find the instantaneous velocity when t = 4.

Learn how to find the average velocity over given time intervals and the instantaneous velocity at a specific time for a particle with displacement s = t^2 - 7t + 15. Includes step-by-step solutions for calculating velocities over intervals such as [3, 4], [3.5, 4], and finding the instantaneous velocity at t = 4.

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