To find the instantaneous velocity of the particle at $t = 0$, we need to calculate the derivative of the displacement function $s(t) = t^2 - t + 22$ with respect to $t$. This derivative, $s'(t)$, will give us the velocity function.

Steps:

1. Differentiate $s(t) = t^2 - t + 22$ with respect to $t$: $s'(t) = \frac{d}{dt}(t^2) - \frac{d}{dt}(t) + \frac{d}{dt}(22)$ Using basic differentiation rules, we get: $s'(t) = 2t - 1 + 0$ So, $s'(t) = 2t - 1$.

2. Substitute $t = 0$ into $s'(t)$ to find the instantaneous velocity at $t = 0$: $s'(0) = 2(0) - 1 = -1$

Answer:

The instantaneous velocity at $t = 0$ is $-1$ kilometers per second.

Would you like more details or have any questions?

Related Questions:

1. How do you find the derivative of higher-degree polynomial functions?
2. What does the sign of the velocity indicate about the particle's direction?
3. How would you find the instantaneous velocity at a different time $t = 1$?
4. How is instantaneous velocity different from average velocity?
5. What is the physical interpretation of the displacement function $s(t)$ in this context?

Tip:

When finding instantaneous rates of change, always differentiate the function first and then substitute the specific value of $t$.