Accumulation Functions and Particle Motion

By the end of this lesson, you will be able to analyze particle motion using integration (displacement, total distance, position from velocity), interpret accumulation functions, and distinguish displacement from total distance traveled.

Accumulation Functions

An accumulation function $F(x) = \int_a^x f(t),dt$ represents the net amount accumulated from $a$ to $x$.

By FTC Part 1: $F'(x) = f(x)$ — the accumulation rate equals the integrand at $x$.

$F$ is increasing where $f > 0$; decreasing where $f < 0$.

Particle Motion Vocabulary

Quantity	Symbol	Meaning	Integral relationship
Position	$s(t)$	Location at time $t$	$s(t) = s(a) + \int_a^t v(u),du$
Velocity	$v(t) = s'(t)$	Rate of position change
Acceleration	$a(t) = v'(t)$	Rate of velocity change	$v(t) = v(a) + \int_a^t a(u),du$
Speed	$	v(t)	$

Displacement vs. Total Distance

Displacement (net change in position):

$$\Delta s = \int_a^b v(t),dt = s(b) - s(a)$$

Positive and negative contributions cancel.

Total distance traveled:

$$D = \int_a^b |v(t)|,dt$$

Split at zeros of $v$ (direction reversals), integrate $|v|$ over each segment.

Example 1 — Position from Velocity

$v(t) = t^2 - 4t + 3$ (m/s), $s(0) = 2$. Find $s(5)$.

$s(t) = s(0) + \int_0^t v(u),du = 2 + \left[\frac{t^3}{3} - 2t^2 + 3t\right]_0^t$

$s(5) = 2 + \frac{125}{3} - 50 + 15 = 2 + \frac{125}{3} - 35 = \frac{125}{3} - 33 = \frac{125 - 99}{3} = \frac{26}{3}$

Example 2 — Displacement vs. Total Distance

$v(t) = t^2 - 4t + 3$ on $[0,5]$.

Zeros of $v$: $t^2-4t+3 = (t-1)(t-3) = 0$ at $t = 1, 3$.

$v > 0$ on $(0,1)$; $v < 0$ on $(1,3)$; $v > 0$ on $(3,5)$.

Displacement: $\int_0^5(t^2-4t+3),dt = \left[\frac{t^3}{3}-2t^2+3t\right]_0^5 = \frac{125}{3}-50+15 = \frac{125-105}{3} = \frac{20}{3}$