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Let's solve the problem step by step. The position of the particle is given by the function:

\[
s(t) = 10t^2 + 20 \, \text{mm}
\]

where \( s \) is the position in millimeters and \( t \) is the time in seconds.

### Part (a): Displacement of the particle from \( t = 1 \, \text{s} \) to \( t = 5 \, \text{s} \)

The displacement is the change in position from \( t = 1 \, \text{s} \) to \( t = 5 \, \text{s} \):

\[
\text{Displacement} = s(5) - s(1)
\]

Calculate \( s(5) \):

\[
s(5) = 10(5)^2 + 20 = 10(25) + 20 = 250 + 20 = 270 \, \text{mm}
\]

Calculate \( s(1) \):

\[
s(1) = 10(1)^2 + 20 = 10(1) + 20 = 10 + 20 = 30 \, \text{mm}
\]

Now, find the displacement:

\[
\text{Displacement} = 270 \, \text{mm} - 30 \, \text{mm} = 240 \, \text{mm}
\]

### Part (b): Average velocity during this time interval

The average velocity \( v_{\text{avg}} \) over the interval from \( t = 1 \, \text{s} \) to \( t = 5 \, \text{s} \) is given by:

\[
v_{\text{avg}} = \frac{\text{Displacement}}{\text{Time interval}} = \frac{s(5) - s(1)}{5 - 1} = \frac{240 \, \text{mm}}{4 \, \text{s}} = 60 \, \text{mm/s}
\]

### Part (c): Acceleration when \( t = 1 \, \text{s} \)

Acceleration is the derivative of velocity with respect to time. First, we need to find the velocity by differentiating the position function with respect to time:

\[
v(t) = \frac{ds(t)}{dt} = \frac{d}{dt}(10t^2 + 20) = 20t \, \text{mm/s}
\]

Next, find the acceleration by differentiating the velocity function with respect to time:

\[
a(t) = \frac{dv(t)}{dt} = \frac{d}{dt}(20t) = 20 \, \text{mm/s}^2
\]

Since the acceleration is constant, the acceleration when \( t = 1 \, \text{s} \) is:

\[
a(1) = 20 \, \text{mm/s}^2
\]

### Summary:
- **(a)** Displacement from \( t = 1 \, \text{s} \) to \( t = 5 \, \text{s} \) is \( 240 \, \text{mm} \).
- **(b)** Average velocity during this interval is \( 60 \, \text{mm/s} \).
- **(c)** Acceleration when \( t = 1 \, \text{s} \) is \( 20 \, \text{mm/s}^2 \).

Would you like more details or have any other questions?

Here are five related questions:
1. How would the displacement change if the interval was \( t = 2 \, \text{s} \) to \( t = 6 \, \text{s} \)?
2. What would the average velocity be over the interval \( t = 0 \, \text{s} \) to \( t = 5 \, \text{s} \)?
3. How would the velocity function change if the position function was \( s(t) = 15t^2 + 30 \)?
4. What is the physical interpretation of constant acceleration in this problem?
5. How would you calculate the instantaneous velocity at any time \( t \)?

**Tip:** For any position function, the first derivative with respect to time gives the velocity, and the second derivative gives the acceleration.

Explore a kinematics problem involving displacement, average velocity, and acceleration. Learn how to calculate displacement, average velocity, and acceleration using the position function s(t) = 10t^2 + 20. Suitable for advanced high school students (Grades 11-12).

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