Motion Along a Line

$v(t) = s'(t)$, $a(t) = v'(t) = s''(t)$

$s(t) = \int v(t)\,dt$, $v(t) = \int a(t)\,dt$

• Speed = $|v(t)|$.
• Particle at rest: $v(t) = 0$.
• Speeding up: $v$ and $a$ have the same sign.
• Slowing down: $v$ and $a$ have opposite signs.
• Displacement: $\int_a^b v(t)\,dt$ (can be negative).
• Total distance: $\int_a^b |v(t)|\,dt$.

$v(t) = t^2 - 4t + 3$.

At rest: $t = 1, t = 3$. $a(t) = 2t - 4$.

At $t = 2$: $v(2) = -1 < 0$, $a(2) = 0$. Neither speeding up nor slowing down.

Displacement $[0,4]$: $\int_0^4 (t^2-4t+3)\,dt = [\frac{t^3}{3}-2t^2+3t]_0^4 = \frac{4}{3}$.