Calculus / Unit 5: Applications of Derivatives I
5.2 Analyzing Motion
Notes
Use the signs of $v$ and $a$.
- The object is at rest when $v = 0$.
- It moves in the positive direction when $v > 0$ and in the negative direction when $v < 0$.
- It is speeding up when $v$ and $a$ have the same sign, and slowing down when they have opposite signs.
- Net displacement is $s(b) - s(a)$. Total distance adds up the distance on each stretch between direction changes.
Practice
A particle has $v = -3$ and $a = -2$. What is happening?
Both are negative, so the speed increases while moving in the negative direction.
For $s(t) = t^3 - 6t^2 + 9t$, when is the particle at rest?
$v = 3t^2 - 12t + 9 = 3(t - 1)(t - 3)$.
A particle has $s(t) = t^2 - 4t$ for $0 \leq t \leq 4$. What total distance does it travel?
$v = 2t - 4$ is 0 at $t = 2$. Positions: $s(0) = 0$, $s(2) = -4$, $s(4) = 0$. Distance is $4 + 4 = 8$, while net displacement is 0.
Ready for more? Take the Unit 5 test.