Calculus / Unit 9: Applications of Integrals
9.5 Motion Problems
Notes
Integrate velocity to get position change.
- Displacement $\displaystyle = \int_a^b v(t)\,dt$.
- Total distance $\displaystyle = \int_a^b |v(t)|\,dt$. Split at times when $v = 0$.
- Final position $\displaystyle = s(a) + \int_a^b v(t)\,dt$.
- Example: $v(t) = 3t^2 - 6t = 3t(t - 2)$ on $[0, 3]$. Displacement is $[t^3 - 3t^2]_0^3 = 0$. The particle moves backward on $(0, 2)$ by 4, then forward on $(2, 3)$ by 4, so the total distance is 8.
Practice
A particle has $v(t) = 3t^2 - 6t$ for $0 \leq t \leq 3$. What is its displacement?
$\displaystyle \int_0^3 (3t^2 - 6t)\,dt = 27 - 27 = 0$.
For the velocity $v(t) = 3t^2 - 6t$ on $[0, 3]$, what is the total distance traveled?
$v < 0$ on $(0, 2)$ giving distance 4, and $v > 0$ on $(2, 3)$ giving distance 4. Total 8.
A particle starts at $s(0) = 5$ with velocity $v(t) = 2t + 1$. What is $s(2)$?
$\displaystyle s(2) = 5 + \int_0^2 (2t + 1)\,dt = 5 + (4 + 2) = 11$.
Ready for more? Take the Unit 9 test.