Problem 4: A particle moves along the x axis and its position as a function of time is x(1) = (17.7 m/s)t - (1.8 m/ 3) ?.
Part (a) What is the instantaneous velocity, in m/s, at t = 2.5 s?
Numeric : A numeric value is expected and not an expression
Part (b) What is the instantaneous velocity, in m/s, at + = 3.9 s?
Numeric
: A numeric value is expected and not an expression.
3
Part (c) What is the average velocity between + = 2.5 s and t = 3.9 s in m/s?

To solve this problem, we need to follow these steps:

### Given:
The position of the particle as a function of time is:
\[ x(t) = (17.7 \ \text{m/s})t - \left(\frac{1.8 \ \text{m}}{3}\right) t^3 \]

We need to find the instantaneous velocity at two specific times and the average velocity between these two times.

### Part (a) - Instantaneous velocity at \( t = 2.5 \ \text{s} \)
The instantaneous velocity is the derivative of the position function with respect to time:
\[ v(t) = \frac{dx(t)}{dt} \]

Given \( x(t) = 17.7t - 0.6t^3 \), differentiate with respect to \( t \):
\[ v(t) = 17.7 - 1.8t^2 \]

Now, plug in \( t = 2.5 \ \text{s} \) to find the instantaneous velocity at that time:
\[ v(2.5) = 17.7 - 1.8(2.5)^2 \]

### Part (b) - Instantaneous velocity at \( t = 3.9 \ \text{s} \)
Similarly, plug in \( t = 3.9 \ \text{s} \) into the velocity function:
\[ v(3.9) = 17.7 - 1.8(3.9)^2 \]

### Part (c) - Average velocity between \( t = 2.5 \ \text{s} \) and \( t = 3.9 \ \text{s} \)
The average velocity is given by the change in position divided by the change in time:
\[ v_{\text{avg}} = \frac{x(3.9) - x(2.5)}{3.9 - 2.5} \]

First, calculate the positions at \( t = 3.9 \ \text{s} \) and \( t = 2.5 \ \text{s} \) using the position function \( x(t) \).

Let’s calculate all the values step by step.### Part (a) - Instantaneous velocity at \( t = 2.5 \ \text{s} \):
The instantaneous velocity at \( t = 2.5 \ \text{s} \) is \( 6.45 \ \text{m/s} \).

### Part (b) - Instantaneous velocity at \( t = 3.9 \ \text{s} \):
The instantaneous velocity at \( t = 3.9 \ \text{s} \) is \( -9.68 \ \text{m/s} \).

### Part (c) - Average velocity between \( t = 2.5 \ \text{s} \) and \( t = 3.9 \ \text{s} \):
The average velocity between \( t = 2.5 \ \text{s} \) and \( t = 3.9 \ \text{s} \) is \( -1.03 \ \text{m/s} \).

If you have any questions or need further details, feel free to ask!

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Here are 5 related questions:
1. What is the acceleration of the particle at \( t = 2.5 \ \text{s} \)?
2. How does the velocity change between \( t = 2.5 \ \text{s} \) and \( t = 3.9 \ \text{s} \)?
3. What is the significance of the negative velocity at \( t = 3.9 \ \text{s} \)?
4. How would the particle's position change if the cubic term had a different coefficient?
5. Can you find the time when the particle comes to rest?

**Tip:** When working with motion problems, always pay attention to the sign of the velocity, as it indicates the direction of motion.

Problem 4: A particle moves along the x axis and its position as a function of time is x(1) = (17.7 m/s)t - (1.8 m/ 3) t^3. Part (a) What is the instantaneous velocity, in m/s, at t = 2.5 s? Part (b) What is the instantaneous velocity, in m/s, at t = 3.9 s? Part (c) What is the average velocity between t = 2.5 s and t = 3.9 s in m/s?

Solve for the instantaneous velocity of a particle at t = 2.5 s and t = 3.9 s, and calculate the average velocity between these two times using kinematic equations. Understand the role of derivatives in determining velocity and motion. Suitable for students in Grades 10-12.

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