Kinematics in One Dimension
Position, Velocity, and Acceleration
For a particle moving along a straight line:
Position: $x(t)$ — the location at time $t$.
Velocity: The rate of change of position:
$$ v(t) = \frac{dx}{dt} $$
Acceleration: The rate of change of velocity:
$$ a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2} $$
Equations of Motion (Constant Acceleration)
When acceleration $a$ is constant:
| Equation | Use |
|---|---|
| $v = v_0 + at$ | Velocity-time, no position |
| $x = x_0 + v_0 t + \frac{1}{2}at^2$ | Position-time |
| $v^2 = v_0^2 + 2a(x - x_0)$ | Velocity-position, no time |
| $x = x_0 + \frac{1}{2}(v_0 + v)t$ | Position, no acceleration |
Free Fall
Under gravity near Earth’s surface:
$$ a = -g \approx -9.8\ \text{m/s}^2 $$
Example Problem
A ball is thrown upward with velocity $20\ \text{m/s}$ from ground level. Find:
1. Maximum height:
$$v^2 = v_0^2 - 2gh$$ $$0 = (20)^2 - 2(9.8)h$$ $$h = \frac{400}{19.6} \approx 20.4\ \text{m}$$
2. Time to reach maximum height:
$$v = v_0 - gt$$ $$0 = 20 - 9.8t$$ $$t \approx 2.04\ \text{s}$$
3. Total time in air:
$$t_{\text{total}} = 2 \times 2.04 \approx 4.08\ \text{s}$$
Relative Velocity
For two objects A and B moving along the same line:
$$ v_{AB} = v_A - v_B $$
where $v_{AB}$ is the velocity of A relative to B.