Particle Motion Using Integration

Particle Motion and Integrals

By the end of this lesson, you will be able to analyze particle motion using both derivatives and integrals of the position function.

The Complete Particle Motion Framework

Quantity	Symbol	From previous	From integration
Position	$s(t)$	Given	$s(t) = \int v(t), dt$
Velocity	$v(t)$	$v(t) = s'(t)$	Given
Acceleration	$a(t)$	$a(t) = v'(t)$	$v(t) = \int a(t), dt$

Finding Position from Velocity

If $v(t)$ is given and we know $s(t_0) = s_0$:

$$s(t) = s_0 + \int_{t_0}^t v(\tau), d\tau$$

This is just the Net Change Theorem: position at time $t$ equals initial position plus displacement.

Example: A particle at position $s(0) = 3$ has velocity $v(t) = 2t - 4$. Find $s(t)$.

$s(t) = 3 + \int_0^t (2\tau - 4), d\tau = 3 + [(\tau^2 - 4\tau)]_0^t = 3 + t^2 - 4t$

$s(t) = t^2 - 4t + 3$

Finding Velocity from Acceleration

$v(t) = v(0) + \int_0^t a(\tau), d\tau$

Displacement vs. Total Distance (Integral Form)

Displacement from $t = a$ to $t = b$:

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

Total distance traveled from $t = a$ to $t = b$:

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

To compute total distance:

1. Find where $v(t) = 0$ on $[a,b]$ (where particle reverses direction)
2. Split the integral at these points
3. Take absolute values of each piece

Example: Complete Analysis

A particle moves with $v(t) = t^2 - 3t + 2 = (t-1)(t-2)$ on $[0, 3]$, $s(0) = 0$.

Position: $s(t) = \int_0^t v, d\tau = \frac{t^3}{3} - \frac{3t^2}{2} + 2t$

Key positions: $s(0) = 0$, $s(1) = \frac{1}{3} - \frac{3}{2} + 2 = \frac{5}{6}$, $s(2) = \frac{8}{3} - 6 + 4 = \frac{2}{3}$, $s(3) = 9 - \frac{27}{2} + 6 = \frac{3}{2}$

$v > 0$ on $(0,1)$; $v < 0$ on $(1,2)$; $v > 0$ on $(2,3)$