https://file.aiquickdraw.com/mathbot/file/17265557104073v6s9ev1.jpg

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.

A particle P is moving on the number line. The displacement s (in cm) of P at time t (in seconds) is given by s = t^2 - 2t. (a) Use the limit definition of the derivative to find the velocity at t = a. (b) Find ds/dt at t = 0, t = 1, t = 3, and determine the motion (left, right, stationary).

This problem involves using the limit definition of the derivative to find the velocity of a particle with displacement s = t^2 - 2t at time t. It covers calculus concepts like the limit definition, derivatives, and velocity at specific times.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation