Question: Suppose the position of a particle, $s(t)$ is given by
$$s(t) = t^2 - 7t + 18$$
feet, where $t$ is time in seconds.
(a) Find the velocity function at time $t$.
$s'(t) = v(t) =$
(b) What is the velocity after 3 seconds with correct units?
(c) Find all value(s) of $t$ (with correct units) for which the particle is at rest (i.e., where $v(t) = 0$). If there are no such values, enter NONE. If there is more than one value, list them separated by commas.
t =
1. **State the problem:** We are given the position function of a particle as $$s(t) = t^2 - 7t + 18$$ where $s(t)$ is in feet and $t$ is in seconds. We need to find the velocity function, the velocity at $t=3$ seconds, and the time(s) when the particle is at rest.
2. **Find the velocity function:** Velocity is the derivative of position with respect to time.
$$v(t) = s'(t) = \frac{d}{dt}(t^2 - 7t + 18)$$
Using the power rule:
$$v(t) = 2t - 7$$
3. **Calculate velocity at $t=3$ seconds:** Substitute $t=3$ into $v(t)$:
$$v(3) = 2(3) - 7 = 6 - 7 = -1$$
The velocity after 3 seconds is $$-1$$ feet per second.
4. **Find when the particle is at rest:** The particle is at rest when velocity is zero.
Set $v(t) = 0$:
$$2t - 7 = 0$$
Solve for $t$:
$$2t = 7$$
$$t = \frac{7}{2}$$
Show cancellation step:
$$t = \frac{\cancel{7}}{\cancel{2}}$$ (no cancellation here, just fraction simplified)
So, the particle is at rest at $$t = 3.5$$ seconds.
**Final answers:**
- Velocity function: $$v(t) = 2t - 7$$
- Velocity at 3 seconds: $$-1$$ feet per second
- Time at rest: $$t = 3.5$$ seconds