Recall:

- \(s(t)\) usually represents a
**position**in one-dimension, either past zero (+) or behind zero (-) - \(v(t) = s'(t)\) usually represents
**velocity**, moving forwards (+) or backwards (-) - \(a(t) = v'(t) = s''(t)\) usually represents
**acceleration**, which can work with velocity (if signs match) or against (if signs are different) - \(|v(t)|\) represents the
**speed**of an object and does NOT have a direction

An object starts by moving at a given velocity and acceleration. Complete the velocities below after the indicated time intervals.

v(t) | a(t) | 1 sec | 2 secs | 3 secs | 4 secs | 5 secs | Speed up/down? |
---|---|---|---|---|---|---|---|

5 m/s | -1 m/s/s | 4 | 3 | 2 | 1 | 0 | down |

0 m/s/s | 5 | 5 | 5 | 5 | 5 | (neither) | |

1 m/s/s | 6 | 7 | 8 | 9 | 10 | up | |

-5 m/s | -1 m/s/s | -6 | -7 | -8 | -9 | -10 | up |

0 m/s/s | -5 | -5 | -5 | -5 | -5 | (neither) | |

1 m/s/s | -4 | -3 | -2 | -1 | 0 | down |

A. An object is **speeding up** when: signs of v(t) and a(t) match

B. An object is **slowing down** when: signs of v(t) and a(t) are opposite

1. Let \( s(t) = t^4 - 8t^3 + 18t^2 \) be the straight-line position of an object at \( t \) (time).

- Find the velocity at \( t=2 \) and \( t=4 \).

\(s'(t) = 4t^3 - 24 t^2 + 36t = 4t (t-3)(t-3) \)

\(s'(2) = 8 \)

\(s'(4) = 16 \) - Find the acceleration at \( t=2 \) and \( t=4 \).

\(s''(t) = 12 t^2 - 48 t + 36 = 12(t-3)(t-1) \)

\(s''(2) = -12 \)

\(s''(4) = 36 \) - Determine when the object is at rest.

\(s'(t) = 4t (t-3)(t-3) = 0 \)

\(t= 0, 3 \) - Determine when the object has no acceleration.

\(s''(t) = 12(t-3)(t-1) = 0 \)

\(t= 1, 3 \) - Over what time intervals is the object speeding up? Slowing down?

\(s''(t) = 12 (t^2 - 4t + 3) = 12 (t-3)(t-1) = 0 \)

\(t= 3, 1 \)

speed up: \( (0, 1) \text{ or } (3,\infty)\)

slow down: \( (1, 3) \)

2. Let \( s(t) = e^t (t-5) \) be the straight-line position of an object at \( t \) (time).

- Find the velocity at \( t=0 \) and \( t=5 \).

\(s'(t) = e^t (t-4) \)

\(s'(0) = -4 \)

\(s'(5) = e^5 \approx 148.41\) - When is the object at rest?

\(s'(t) = e^t (t-4) = 0 \)

\(t = 4 \) [note: e^t is never zero] - When is the object moving backwards?

interval \([0,4)\) - When is the object speeding up?

\( s''(t) = e^t (t - 3) = 0 \)

\( t = 3 \)

speeding up: \( [0, 3) \) or \( (4, \infty) \)

Example: If I go forward 10 miles, backwards 3 miles, then forward 5 miles, my **distance** traveled is 18 miles. However, my **displacement** is 12 miles from where I started. Distance is always positive.

Let \(s(t) = (t-5)(t-10) = t^2 - 15t + 50 \) be the straight-line position of an object at time \(t\).

- When does the object change directions?

\(s'(t) = 2t - 15 = 0 \)

\( t = \frac{15}{2} = 7.5 \) - Find the total
**distance**traveled on the interval \( [0, 10] \). Hint: How far did the object move backward? How far did the object move forward?

backward: \(| s(7.5) - s(0) | = |-6.25 - 50| = 56.25 \)

forward: \(s(10) - s(7.5) | ~= |0 - 6.25| = 6.25 \)

\(56.25 + 6.25 ~= 62.5 \) - What is the
**displacement**of the object over the interval \( [0, 10] \)?

\(s(10) - s(0) = 0 - 50 = -50 \)