The problem describes the motion of an object along the x-axis, where the position $s(t)$ is given by the function:

$s(t) = t^3 - 6t^2 - 15t$

We need to analyze the motion at $t = 1$ second. The given results for position, velocity, and acceleration at $t = 1$ are:

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

Here’s how these values are calculated step-by-step:

1. Position at $t = 1$

$s(t) = t^3 - 6t^2 - 15t$ Substitute $t = 1$: $s(1) = (1)^3 - 6(1)^2 - 15(1)$ $s(1) = 1 - 6 - 15 = -20 \, \text{feet}.$

2. Velocity at $t = 1$

The velocity is the first derivative of the position function: $v(t) = s'(t) = 3t^2 - 12t - 15$ Substitute $t = 1$: $v(1) = 3(1)^2 - 12(1) - 15$ $v(1) = 3 - 12 - 15 = -24 \, \text{ft/s}.$

3. Acceleration at $t = 1$

The acceleration is the derivative of the velocity function (or the second derivative of the position function): $a(t) = v'(t) = s''(t) = 6t - 12$ Substitute $t = 1$: $a(1) = 6(1) - 12$ $a(1) = 6 - 12 = -6 \, \text{ft/s}^2.$

Summary of Results

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

Questions for Exploration:

1. How would the motion change if the initial position at $t = 0$ were altered?
2. What does a negative velocity at $t = 1$ indicate about the direction of the motion?
3. How would the object’s motion differ if the acceleration were positive instead of negative?
4. Can you find the time $t$ when the object reaches maximum or minimum velocity?
5. How does the total displacement differ from the distance traveled over the interval $t \in [0, 1]$?

Tip:

When analyzing motion problems, always find the critical points of velocity and acceleration to understand the object’s behavior over time.