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.