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Let's go step by step to solve the given problem.

The displacement function \( s(t) \) of a particle is given by:

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

### (a) Find the average velocity over each time interval:

The average velocity over a time interval \([t_1, t_2]\) is given by:

\[
v_{avg} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}
\]

Let’s compute the average velocity for each sub-interval.

#### (i) For the interval \([3, 4]\):

- \( s(3) = 3^2 - 7(3) + 15 = 9 - 21 + 15 = 3 \)
- \( s(4) = 4^2 - 7(4) + 15 = 16 - 28 + 15 = 3 \)

Now calculate the average velocity:

\[
v_{avg} = \frac{s(4) - s(3)}{4 - 3} = \frac{3 - 3}{1} = 0 \, \text{m/s}
\]

#### (ii) For the interval \([3.5, 4]\):

- \( s(3.5) = (3.5)^2 - 7(3.5) + 15 = 12.25 - 24.5 + 15 = 2.75 \)
- \( s(4) = 3 \) (as computed earlier)

Now calculate the average velocity:

\[
v_{avg} = \frac{s(4) - s(3.5)}{4 - 3.5} = \frac{3 - 2.75}{0.5} = 0.5 \, \text{m/s}
\]

#### (iii) For the interval \([4, 5]\):

- \( s(4) = 3 \) (as computed earlier)
- \( s(5) = 5^2 - 7(5) + 15 = 25 - 35 + 15 = 5 \)

Now calculate the average velocity:

\[
v_{avg} = \frac{s(5) - s(4)}{5 - 4} = \frac{5 - 3}{1} = 2 \, \text{m/s}
\]

#### (iv) For the interval \([4, 4.5]\):

- \( s(4) = 3 \) (as computed earlier)
- \( s(4.5) = (4.5)^2 - 7(4.5) + 15 = 20.25 - 31.5 + 15 = 3.75 \)

Now calculate the average velocity:

\[
v_{avg} = \frac{s(4.5) - s(4)}{4.5 - 4} = \frac{3.75 - 3}{0.5} = 1.5 \, \text{m/s}
\]

### (b) Find the instantaneous velocity when \( t = 4 \).

The instantaneous velocity is the derivative of the displacement function \( s(t) \), which gives the velocity function \( v(t) \):

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

Now, plug in \( t = 4 \):

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

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### Summary of answers:

(a) Average velocities:
1. \( [3, 4] \): \( 0 \, \text{m/s} \)
2. \( [3.5, 4] \): \( 0.5 \, \text{m/s} \)
3. \( [4, 5] \): \( 2 \, \text{m/s} \)
4. \( [4, 4.5] \): \( 1.5 \, \text{m/s} \)

(b) Instantaneous velocity at \( t = 4 \): \( 1 \, \text{m/s} \)

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Let me know if you need further details on any part, or have any questions!

### Related Questions:
1. What is the meaning of average velocity over an interval?
2. How do you differentiate a quadratic function?
3. What is the geometric interpretation of instantaneous velocity?
4. How does the velocity change over time for this particle?
5. How do you calculate displacement from a velocity function?

**Tip:** When working with displacement or velocity functions, remember that the instantaneous velocity is the derivative of the displacement, and acceleration is the derivative of the velocity!

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 calculate the average velocity over given intervals and the instantaneous velocity of a particle with a displacement function s = t^2 - 7t + 15. This problem covers concepts of calculus, such as derivatives, average velocity, and instantaneous velocity, suitable for students in Grades 11-12.

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