The question provides the position function $s(t) = t^3 - 6t^2 - 15t$ and asks to analyze the motion at $t = 1$ second.

Analysis:

1. Position ($s(t)$): The position function is $s(t) = t^3 - 6t^2 - 15t$. At $t = 1$: $s(1) = (1)^3 - 6(1)^2 - 15(1) = 1 - 6 - 15 = -20 \, \text{feet.}$

2. Velocity ($v(t)$): The velocity is the first derivative of the position function: $v(t) = \frac{ds}{dt} = 3t^2 - 12t - 15.$ At $t = 1$: $v(1) = 3(1)^2 - 12(1) - 15 = 3 - 12 - 15 = -24 \, \text{ft/s.}$

3. Acceleration ($a(t)$): The acceleration is the derivative of the velocity function: $a(t) = \frac{dv}{dt} = 6t - 12.$ At $t = 1$: $a(1) = 6(1) - 12 = -6 \, \text{ft/s}^2.$

Summary of Results:

• $s(1) = -20 \, \text{feet}$,
• $v(1) = -24 \, \text{ft/s}$,
• $a(1) = -6 \, \text{ft/s}^2$.

Explanation of Motion:

1. The position is negative, meaning the object is 20 feet behind the origin.
2. The velocity is negative, indicating the object is moving to the left along the x-axis at $24 \, \text{ft/s}$.
3. The acceleration is negative, meaning the object is experiencing a deceleration in the leftward direction.

Would you like further explanations on these calculations? Let me know!

Related Questions:

1. How does the velocity change with respect to time for this function?
2. What is the object's acceleration at $t = 0$ or $t = 2$?
3. At what time(s) does the object come to rest ($v(t) = 0$)?
4. How would you determine if the motion is speeding up or slowing down?
5. What is the significance of a negative position, velocity, and acceleration in this context?

Tip:

The sign of the velocity and acceleration determines whether an object is speeding up or slowing down. When they have the same sign, the object speeds up; when they differ, it slows down.