Let's solve this step by step.

Part (a): Finding the velocity using the limit definition of the derivative

The displacement of the particle is given by the function:

$s(t) = t^2 - 2t$

To find the velocity at time $t = a$, we use the limit definition of the derivative:

$v(a) = \lim_{h \to 0} \frac{s(a+h) - s(a)}{h}$

1. Find $s(a+h)$:

$s(a+h) = (a+h)^2 - 2(a+h)$ Expanding this:

$s(a+h) = a^2 + 2ah + h^2 - 2a - 2h$

2. Substitute $s(a+h)$ and $s(a)$ into the limit definition:

Since $s(a) = a^2 - 2a$, we have:

$v(a) = \lim_{h \to 0} \frac{(a^2 + 2ah + h^2 - 2a - 2h) - (a^2 - 2a)}{h}$

Simplifying the expression inside the numerator:

$= \lim_{h \to 0} \frac{a^2 + 2ah + h^2 - 2a - 2h - a^2 + 2a}{h}$

Notice that $a^2$ and $-a^2$, as well as $-2a$ and $2a$, cancel out, leaving:

$v(a) = \lim_{h \to 0} \frac{2ah + h^2 - 2h}{h}$

Factor out $h$ from the numerator:

$v(a) = \lim_{h \to 0} \frac{h(2a + h - 2)}{h}$

Cancel the $h$:

$v(a) = \lim_{h \to 0} (2a + h - 2)$

Now, as $h \to 0$, the expression becomes:

$v(a) = 2a - 2$

So, the velocity function is:

$v(t) = 2t - 2 \quad \text{(in centimeters per second, cm/s)}$

Part (b): Finding $\frac{ds}{dt}$ at specific times

We already found that $\frac{ds}{dt} = v(t) = 2t - 2$.

• At $t = 0$:

$v(0) = 2(0) - 2 = -2 \, \text{cm/s}$ (The particle is moving to the left since the velocity is negative.)

• At $t = 1$:

$v(1) = 2(1) - 2 = 0 \, \text{cm/s}$ (The particle is stationary since the velocity is zero.)

• At $t = 3$:

$v(3) = 2(3) - 2 = 6 - 2 = 4 \, \text{cm/s}$ (The particle is moving to the right since the velocity is positive.)

Summary of the results:

• At $t = 0$, the particle is moving to the left with a velocity of $-2 \, \text{cm/s}$.
• At $t = 1$, the particle is stationary.
• At $t = 3$, the particle is moving to the right with a velocity of $4 \, \text{cm/s}$.

Would you like further details on any of these steps?

Here are 5 related questions to expand on this topic:

1. How does the limit definition of a derivative relate to instantaneous velocity?
2. What happens to the motion of the particle as $t \to \infty$?
3. How would the velocity change if the displacement function were different, say $s(t) = t^3 - 2t$?
4. What is the significance of the points where velocity equals zero?
5. How can you find the acceleration of the particle from the velocity function?

Tip: When interpreting a velocity-time graph, remember that negative velocity indicates motion in the opposite direction from positive velocity.