To solve the given problem, we start by recognizing that the displacement $s(t)$ is given by the equation:

$s(t) = t^2 - 6t + 19$

(A) Average Velocity

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 compute the average velocity for each of the specified time intervals:

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(i) Average velocity over the interval [3, 4]:

We need to find $s(3)$ and $s(4)$:

$s(3) = 3^2 - 6(3) + 19 = 9 - 18 + 19 = 10$ $s(4) = 4^2 - 6(4) + 19 = 16 - 24 + 19 = 11$

Now, the average velocity is:

$\frac{s(4) - s(3)}{4 - 3} = \frac{11 - 10}{1} = 1 \, \text{m/s}$

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(ii) Average velocity over the interval [3.5, 4]:

We need to find $s(3.5)$ and $s(4)$:

$s(3.5) = (3.5)^2 - 6(3.5) + 19 = 12.25 - 21 + 19 = 10.25$ $s(4) = 11 \, \text{(from previous calculation)}$

Now, the average velocity is:

$\frac{s(4) - s(3.5)}{4 - 3.5} = \frac{11 - 10.25}{0.5} = \frac{0.75}{0.5} = 1.5 \, \text{m/s}$

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(iii) Average velocity over the interval [4, 5]:

We need to find $s(5)$ and $s(4)$:

$s(5) = 5^2 - 6(5) + 19 = 25 - 30 + 19 = 14$ $s(4) = 11 \, \text{(from previous calculation)}$

Now, the average velocity is:

$\frac{s(5) - s(4)}{5 - 4} = \frac{14 - 11}{1} = 3 \, \text{m/s}$

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(iv) Average velocity over the interval [4, 4.5]:

We need to find $s(4.5)$ and $s(4)$:

$s(4.5) = (4.5)^2 - 6(4.5) + 19 = 20.25 - 27 + 19 = 12.25$ $s(4) = 11 \, \text{(from previous calculation)}$

Now, the average velocity is:

$\frac{s(4.5) - s(4)}{4.5 - 4} = \frac{12.25 - 11}{0.5} = \frac{1.25}{0.5} = 2.5 \, \text{m/s}$

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(B) Instantaneous Velocity at $t = 4$

The instantaneous velocity is given by the derivative of $s(t)$ with respect to $t$:

$v(t) = \frac{ds}{dt} = 2t - 6$

Now, substitute $t = 4$:

$v(4) = 2(4) - 6 = 8 - 6 = 2 \, \text{m/s}$

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Final Results:

• (i) Average velocity over [3,4]: $1 \, \text{m/s}$
• (ii) Average velocity over [3.5,4]: $1.5 \, \text{m/s}$
• (iii) Average velocity over [4,5]: $3 \, \text{m/s}$
• (iv) Average velocity over [4,4.5]: $2.5 \, \text{m/s}$
• (B) Instantaneous velocity at $t = 4$: $2 \, \text{m/s}$

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Would you like further details or have any questions?

Here are 5 related questions to explore:

1. What does it mean when the average and instantaneous velocities are equal?
2. How can the direction of the particle’s motion be inferred from velocity?
3. How would you find the time when the particle is at rest?
4. What is the significance of negative velocity in this context?
5. Can you graph the displacement and velocity functions?

Tip:

The instantaneous velocity is the slope of the tangent line to the displacement curve at a given point.