notes/education/physics/PHYS2210/Unit 1.md

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

Kinematics problems have a start and an end of the motion.

Initial	Final
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?