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.