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.
\(s'(t) = 4t (t-3)(t-3) = 0 \)
\(t = 0, t =3 \)
Determine when the object has no acceleration.
\(s''(t) = 12(t-3)(t-1) = 0 \)
\(t = 1, t = 3 \)
Over what time intervals is the object speeding up? Slowing down?
speed up: \( (0, 1) \cup (3,\infty)\)
slow down: \( (1, 3) \)
2. Let \( s(t) = e^t (t-5) \) for \( t \ge 0 \), 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: \( [0, 3) \cup (4, \infty) \)
Challenge: Distance vs Displacement
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 \( [0, 10] \)?
\(s(10) - s(0) = 0 - 50 = -50 \)