Physics makes models that approximate reality with math. Physics models have made amazing progress, but so far no model perfectly matches reality.

x y Vx Vy

This model approximates particle motion and collisions. It gives each particle a position and velocity. Every second, a particle's velocity is added to its position. Also, since this is a 2-D model, the calculations are repeated in the horizontal and vertical.

Question: Imagine this is a simulation of billiards or pool. What does this simulation need to be more realistic?

  • there could be pockets
  • balls that are the same size
  • balls that can roll
  • air friction
  • rolling friction
  • when the balls collide they make a sound
  • collisions that lose energy
  • a 3rd dimension, so the balls can bounce
  • gravity
  • walls that aren't perfectly rigid rectangles

    At the extremes we'd also need things like:
  • slightly bumpy shaped balls, not perfect spheres
  • quantum effects, like tunneling
  • effects of relativity, like time dilation

  • Position

    We use the symbol x for position with meters for units. You might be used to using x in other ways, but in physics it is reserved for position.

    To represent a displacement we subtract the initial position from the final position. We use the Δ symbol to show change.

    $$ \Delta x = x_f-x_i$$

    \(\Delta x\) = change in position, displacement [m, meters] vector

    \(x_i\) = initial position [m, meters] vector

    \(x_f\) = final position [m, meters] vector

    Δx=0 Δt=0 v=0 a=0
    Example: Mount Everest is 8848 m above sea level. The base camp, where most mountain climbers begin their trek, is 5150 m above sea level. What is the vertical distance between the base camp and the peak?
    solution $$ \text{final position} = 8848\,\mathrm{m} $$ $$\text{initial position} = 5150\,\mathrm{m}$$
    $$ \Delta x = x_{f}-x_{i}$$ $$ \Delta x = 8848\,\mathrm{m}-5150\,\mathrm{m}$$ $$ \Delta x = 3698\,\mathrm{m}$$
    Example: Convert your answer from the previous problem into miles and kilometers. Which is easier to convert?
    1 mile = 1609 meters
    1 kilometer = 1000 meters
    solution $$ \Delta x = 3698\,\mathrm{m}$$
    $$ 3698\,\mathrm{m} \left( \mathrm{ \frac{1 \, mile}{1609.34 \, m}} \right) = 2.297 \, \mathrm{mile} $$
    $$3698\,\mathrm{m} = 3\overleftarrow{\undergroup{6}\undergroup{9}\undergroup{8}}. \, \mathrm{m}= 3.698 \,\mathrm{km}$$
    Distance traveled refers to the length of a path.

    Displacement is a change in position. The path taken is irrelevant.

    My school is only 1.5 km from where I live. I'm able to ride my bike most days. One morning I had biked half a kilometer before I realized I forgot my lunch. So, I rode home, grabbed my lunch, and rushed to school.

    Example: What's my displacement?

    Displacement is final position minus initial position. Heading backwards, going slower, or taking a different path doesn't change its value.

    $$ \Delta x = x_f-x_i$$ $$ \Delta x = 1.5 \, \mathrm{km}-0$$ $$ \Delta x = 1.5 \, \mathrm{km}$$
    Example: What's my distance traveled?

    Distance traveled measures the length of the path. It will include heading back home to get my lunch.

    $$d = 0.5 \, \mathrm{km} + 0.5 \, \mathrm{km} + 1.5 \, \mathrm{km}$$ $$d = 2.5 \, \mathrm{km}$$


    We keep track of time in the same way as position. We use t as the symbol with seconds as our main unit.

    $$\Delta t = t_{f}-t_{i}$$

    \(\Delta t\) = change in time, time period [s, seconds]

    \(t_i\) = initial time [s, seconds]

    \(t_f\) = final time [s, seconds]

    Δx=0 Δt=0 v=0 a=0
    Example: School ends at 2:30 PM. I got home today at 2:49 PM. How long did it take me to get home in minutes?

    We can just focus on the minutes, since the hours didn't change.

    $$\Delta t = t_{f}-t_{i}$$ $$\Delta t = 49\,\mathrm{min}-30\, \mathrm{min}$$ $$\Delta t = 19\, \mathrm{min}$$

    Example: My school starts early at 7:30 am. Each class is 1 hour long with 3 minutes between classes. What time is it at the end of the second period?

    You can probably solve this in your head, but try using Δt = tf - ti for practice.
    solution $$\Delta t = 1\, \mathrm{hr} + 1 \, \mathrm{hr} + 3\, \mathrm{min}$$ $$\Delta t = 2\, \mathrm{hr} + 3\, \mathrm{min}$$
    $$\Delta t = t_{f}-t_{i}$$ $$t_{f} = \Delta t + t_{i}$$ $$t_{f} = (2\, \mathrm{hr} + 3\, \mathrm{min}) + (7\, \mathrm{hr} + 30\, \mathrm{min})$$ $$t_{f} = 9\, \mathrm{hr} + 33\, \mathrm{min}$$


    Velocity measures how much position changes, Δx, over a period of time, Δt.

    $$v_{\mathrm{avg}} = \frac{\Delta x}{\Delta t}$$

    \(\Delta x\) = change in position, displacement [m, meters] vector

    \(\Delta t\) = time period [s, seconds]

    \(v\) = average velocity [m/s, meters per second] vector

    Δx=0 Δt=0 v=0 a=0
    Example: A car is traveling at an average of 2 m/s for 80 seconds. How far does the car travel?
    solution $$ v = \frac{\Delta x}{\Delta t} $$ $$ v \Delta t= \Delta x$$ $$ (2 \mathrm{\tfrac{m}{s}}) (80\, \mathrm{s}) = \Delta x $$ $$ (2 \mathrm{\tfrac{m}{\cancel{s}}}) (80\, \mathrm{\cancel{s}}) = \Delta x $$ $$ 160\, \mathrm{m} = \Delta x $$ Δx=0 Δt=0 v=0 a=0
    Question: Why are we solving for average velocity? Why not just velocity?

    The velocity might change over the journey. What if someone speeds up or slows down? There are many possible velocities that fit a given distance and time, but only one average velocity.

    A B 5 10 152025303540455055510152025 time (s) position (m)

    In the next unit we will learn how to find the instantaneous velocity.

    Example: Someone tells you they can run 10 km in about an hour. What velocity is that in m/s? Is that fast? (1 m/s is walking speed)
    solution $$10 \, \mathrm{km} = 10(1000) \mathrm{m} = \color{#f07} 10\,000 \, \mathrm{m}$$ $$1\, \mathrm{h} \left(\frac{60\, \mathrm{min}}{1\, \mathrm{h}}\right)\left(\frac{60\, \mathrm{s}}{1\, \mathrm{min}}\right) = \color{#09d} 3600 \,\mathrm{s}$$
    $$v = \frac{\Delta x}{\Delta t}$$ $$v = \frac{\color{#f07}10\,000\, \mathrm{m}}{\color{#09d} 3600\, \mathrm{s}}$$ $$v = 2.7\, \mathrm{\tfrac{m}{s}}$$
    Example: Google maps says Las Vegas is 4 hours away from Los Angeles. Google says it is 270 miles away. How fast does google think I will drive? Answer this one in miles/hour.
    solution $$v = \frac{\Delta x}{\Delta t}$$ $$v = \frac{270\, \mathrm{miles} }{4\, \mathrm{hour}}$$ $$v = 67.5\, \mathrm{\tfrac{miles}{hour} }$$

    The posted speed limit is 70 mph.

    Question: Imagine you are in a plane with your eyes closed. Can you feel when the plane is increasing or decreasing altitude? What does it feel like?

    A plane dropping at a constant 100 m/s would feel the same as a plane staying at the same altitude. There is no way to directly measure your velocity relative to the Earth, but you can indirectly guess your velocity by looking outside, or measuring air pressure.

    You can feel acceleration, the change in velocity. For me, accelerating down feels like butterflies, and accelerating up feels like pressure.

    Example: If I walk at a speed of 1.2 m/s how long will it take for me to walk 2 km?
    solution $$2 \, \mathrm{km} = 2(1000) \mathrm{m} = 2000 \, \mathrm{m}$$
    $$v = \frac{\Delta x}{\Delta t}$$ $$\Delta t = \frac{\Delta x}{v}$$ $$\Delta t = \frac{2000\, \mathrm{m}}{1.2\, \mathrm{\tfrac{m}{s}}}$$ $$\Delta t = \frac{2000}{1.2}\, \mathrm{s}$$ $$\Delta t = 1666.\overline{6} \, \mathrm{s}$$
    Question: This graph shows position vs. time. What does its slope represent?


    Velocity is defined as the change in position every second. This is the same as the rise over the run, which is slope.


    Acceleration is a measure of how much velocity changes (Δv) over a period of time (Δt).

    $$a_{\mathrm{avg}} = \frac{\Delta v}{\Delta t}$$

    \(\Delta v\) = change in velocity [m/s] = \(v_f-v_i\) vector

    \(\Delta t\) = time period, change in time [s, seconds]

    \(a\) = acceleration [m/s²] vector

    Δx=0 Δt=0 v=0 a=0
    Example: A basketball falls off a table and hits the floor in 0.45 s. The ball has a velocity of 4.43 m/s right before it hits the ground. What is the acceleration of the basketball as it falls?
    solution $$a = \frac{\Delta v}{\Delta t}$$ $$a = \frac{4.43\, \mathrm{\tfrac{m}{s}}} {0.45\, \mathrm{s}}$$ $$a = 9.84\, \mathrm{\tfrac{m}{s^{2}} }$$
    Example: A Tesla can go from 0 to 60 miles per hour in 2.28 s. What is the acceleration in m/s²? (1 mile = 1609 meters)
    solution $$ 60\left( \mathrm{ \frac{\color{#f05}{mile}}{\color{#07b}{hour}}} \right)\left(\frac{1609\,\mathrm{ m}}{1 \,\color{#f05}{\mathrm{mile} }}\right)\left(\frac{1\, \color{#07b}{ \mathrm{hour} }}{3600\, \mathrm{s} }\right) = 26.8 \mathrm{\tfrac{m}{s}} $$ $$a = \frac{\Delta v}{\Delta t}$$ $$a = \frac{26.8\,\mathrm{ \tfrac{m}{s}}} {2.28\, \mathrm{s}}$$ $$a = 11.8\, \mathrm{ \tfrac{m}{s^{2}} }$$ Δx=0 Δt=0 v=0 a=0
    Question: People often get velocity and acceleration confused. Explain the difference between them without math.

    Velocity is how fast you are going. It's how quick you change your position. Velocity doesn't feel like anything, but it looks like stuff going past you.

    Acceleration is how fast your speed changes. It's your change in velocity. You can feel acceleration in your body. It feels like being pushed. If you are falling it feels like butterflies.

    Example: I start a velocity of 1 m/s. I speed up to 3 m/s over 10 seconds. What is my acceleration?
    $$\Delta v = v_{f}-v_{i}$$ $$\Delta v = 3 \,\mathrm{ \tfrac{m}{s}} -1\, \mathrm{ \tfrac{m}{s} }$$ $$\Delta v = 2\, \mathrm{\tfrac{m}{s}}$$
    $$a = \frac{\Delta v}{\Delta t}$$ $$a = \frac{2\, \mathrm{\tfrac{m}{s}} }{10\, \mathrm{s} }$$ $$a = 0.2\,\mathrm{ \tfrac{m}{s^{2}}}$$
    Question: This graph shows velocity vs. time. What does its slope represent?


    Acceleration is defined as the change in velocity every second. This is the same as the rise over the run, which is slope.

    Graphing Motion

    Velocity is the slope of the position vs. time graph.
    Acceleration is the slope of the velocity vs. time graph.

    If the acceleration is positive, the velocity increases with time.
    If the velocity is positive the position increases with time.

    acceleration = m/s²
    Graph: Use this graph of velocity vs. time to determine acceleration.

    Acceleration is the slope of the velocity vs. time graph. Slope is the vertical rise of the graph divided by the horizontal run.

    $$\mathrm{slope} = \frac{\mathrm{rise}}{\mathrm{run}}$$ $$a = \frac{\Delta v}{\Delta t}$$ $$a = \frac{1}{4}$$
    Graph: Sketch the position vs. time and velocity vs. time graphs for an object falling for 10 s. For your sketch, just make sure your graphs have the right shape.
    The acceleration is -9.8 m/s².
    The magnitude of the velocity will increase by about 10 m/s every second.
    The direction of the velocity is negative so the velocity should slope down
    The position should also curve downwards.

    Graph: Draw the position vs. time and velocity vs. time graphs for an object moving at 2 m/s for 10 s.

    Acceleration doesn't have to be constant. In the graphs below, acceleration is increasing. The change in acceleration over time is called jerk.

    jerk = m/s³

    A large jerk can cause you to stumble. A jerk could come from a short lasting push or pull. It could come from a car starting to break, or accelerate. Falling causes a jerk. Jumping. Bouncing. Any change in acceleration.