Motion in a straight line is one dimensional.

Kinematics or the physics of motion has 4 noteworthy parameters: time ($t$), position ($x$), velocity ($v$), and acceleration ($a$). 

Kinematic problems have a start and end of motion.

# Displacement

Displacement is calculated with the formula:
$$\Delta x = \text{x-value of final position} - \text{x-value of initial posiion}$$

# Velocity
Average velocity over a time interval $\Delta t$ is defined to be: **the displacement** (net change in position), **divided by** **the time taken**.

$$ \bar{v} = \dfrac{\text{final position-initial position}}{\text{final time - initial time}} = \dfrac{x_2 - x_1}{t_2 - t_1} = \dfrac{\Delta x}{\Delta t}$$
**Speed (m/s)** is defined to be the total distance traveled divided by the time taken. Speed and velocity are *not the same*. 


$$ v_{\text{instant}} = v = \lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t} = \frac{dx}{dt}$$
- $x(t)$ -> **position** as a function of time
- $v(t)$ -> **slope** of position-vs-time (derivative of $x(t)$)
- $a(t)$ -> **slope** of velocity-vs-time (derivative of $v(t)$)
# Acceleration
To find the instantaneous acceleration, we can apply the formula:

$$a_{\text{instant}} = a = \frac{dv}{dt} = \frac{d}{dt} \frac{dx}{dt} = \frac{d^2x}{dt^2}$$
## Equations of Motion for Constant Acceleration
1. $v = v_0 + at$  - Use when missing $x$
2. $x = x_0 + \frac{1}{2}(v_0 + v)t$  - Use when missing $a$
3. $x = x_0 + v_0 t + \frac{1}{2} a t^2$  - Use when missing $v$
4. $v^2 = v_0^2 + 2a(x - x_0)$  - Use when missing $t$ 

(Actually useful equations)
$$ t = \frac{v_f - v_i}{a} $$
$$s = v_it + \frac{1}{2}at^2 $$
| -------------- | ----- |
| $t_0$          | $t$   |
| $v_0$          | $v$   |
| $x_0$          | $x$   |
| $a$ (constant) | $a$   |
## Examples

 > Sally aggressively drives her Alfa Romeo from rest to 50 m/s in 6s. What is her acceleration?
1. Start:
	- $v_0 = 0 m/s$
	- $t_0 = 0s$
	- $x_0 = 0m$
2. Final:
	- $v = 50 m/s$
	- $t = 6s$
	- $x = ?$ 
3. Using the equation $v = v_0 + at \to a = \frac{(v - v_0)}{t}$, $a$ = $(50 m/s - 0 m/s)/ 6s = 8.3m/s^2$ 
4. Assess: do the units make sense? Is the answer reasonable?

> An object slides until coming to a rest. It traveled 15 meters in 3 seconds. What was the object's acceleration?
1. The problem is a kinematics problem, refer to the above formula.
2. Acceleration is a constant
3. Problem has two unknowns, $v_0$ and $a$, let's solve for $v_0$ first.
4. $x = x_0 + \frac{1}{2}(v_0 + v) t \to v_0 = 2*(x - x_0)/ t \to 2*(15)/3 = 10$
5. $v = v_0 + at \to a = (v - v_0) /t \to a = (-10 m/s) / 3s = -3.33m/s^2$ 
6. Asses: Units? Answer make sense? Significant figures? Acceleration speeding up or slowing down?
# Gravity
- Is the problem 1 dimensional?
- Could draw axis pointing upwards, call it y axis
- If not specified, assume acceleration is $g = 9.8 m/s^2$.
- Direction is important, $g$ is down towards the earth, so it's often negative. The sign depends on your choice of axis.