Physics makes models that approximate reality with math. Physics models have made amazing progress, but so far no model perfectly matches reality.
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.
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
solution
$$ \text{final position} = 8848\,\mathrm{m} $$ $$\text{initial position} = 5150\,\mathrm{m}$$$$ \Delta x = x_{f}-x_{i}$$ $$ \Delta x = 8848\,\mathrm{m}-5150\,\mathrm{m}$$ $$ \Delta x = 3698\,\mathrm{m}$$
1 mile = 1609 meters
1 kilometer = 1000 meters
solution
$$ \Delta x = 3698\,\mathrm{m}$$$$ 3698\,\mathrm{m} \left( \mathrm{ \frac{1 \, mile}{1609.34 \, m}} \right) = 2.297 \, \mathrm{mile} $$
$$3698\,\mathrm{m} = 3\overleftarrow{\undergroup{6}\undergroup{9}\undergroup{8}}. \, \mathrm{m}= 3.698 \,\mathrm{km}$$
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.
solution
Displacement is final position minus initial position. Heading backwards, going slower, or taking a different path doesn't change its value.
$$ \Delta x = x_f-x_i$$ $$ \Delta x = 1.5 \, \mathrm{km}-0$$ $$ \Delta x = 1.5 \, \mathrm{km}$$solution
Distance traveled measures the length of the path. It will include heading back home to get my lunch.
$$d = 0.5 \, \mathrm{km} + 0.5 \, \mathrm{km} + 1.5 \, \mathrm{km}$$ $$d = 2.5 \, \mathrm{km}$$Time
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]
solution
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
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
solution
$$ v = \frac{\Delta x}{\Delta t} $$ $$ v \Delta t= \Delta x$$ $$ (2 \mathrm{\tfrac{m}{s}}) (80\, \mathrm{s}) = \Delta x $$ $$ (2 \mathrm{\tfrac{m}{\cancel{s}}}) (80\, \mathrm{\cancel{s}}) = \Delta x $$ $$ 160\, \mathrm{m} = \Delta x $$answer
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.
In the next unit we will learn how to find the instantaneous velocity.
solution
$$10 \, \mathrm{km} = 10(1000) \mathrm{m} = \color{#f07} 10\,000 \, \mathrm{m}$$ $$1\, \mathrm{h} \left(\frac{60\, \mathrm{min}}{1\, \mathrm{h}}\right)\left(\frac{60\, \mathrm{s}}{1\, \mathrm{min}}\right) = \color{#09d} 3600 \,\mathrm{s}$$$$v = \frac{\Delta x}{\Delta t}$$ $$v = \frac{\color{#f07}10\,000\, \mathrm{m}}{\color{#09d} 3600\, \mathrm{s}}$$ $$v = 2.7\, \mathrm{\tfrac{m}{s}}$$
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.
answer
A place 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.
solution
$$2 \, \mathrm{km} = 2(1000) \mathrm{m} = 2000 \, \mathrm{m}$$$$v = \frac{\Delta x}{\Delta t}$$ $$\Delta t = \frac{\Delta x}{v}$$ $$\Delta t = \frac{2000\, \mathrm{m}}{1.2\, \mathrm{\tfrac{m}{s}}}$$ $$\Delta t = \frac{2000}{1.2}\, \mathrm{s}$$ $$\Delta t = 1666.\overline{6} \, \mathrm{s}$$
answer
velocity
Velocity is defined as the change in position every second. This is the same as the rise over the run, which is slope.
Acceleration
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
solution
$$a = \frac{\Delta v}{\Delta t}$$ $$a = \frac{4.43\, \mathrm{\tfrac{m}{s}}} {0.45\, \mathrm{s}}$$ $$a = 9.84\, \mathrm{\tfrac{m}{s^{2}} }$$solution
$$ 60\left( \mathrm{ \frac{\color{#f05}{mile}}{\color{#07b}{hour}}} \right)\left(\frac{1609\,\mathrm{ m}}{1 \,\color{#f05}{\mathrm{mile} }}\right)\left(\frac{1\, \color{#07b}{ \mathrm{hour} }}{3600\, \mathrm{s} }\right) = 26.8 \mathrm{\tfrac{m}{s}} $$ $$a = \frac{\Delta v}{\Delta t}$$ $$a = \frac{26.8\,\mathrm{ \tfrac{m}{s}}} {2.28\, \mathrm{s}}$$ $$a = 11.8\, \mathrm{ \tfrac{m}{s^{2}} }$$solution
answer
acceleration
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.
solution
Acceleration is the slope of the velocity vs. time graph. Slope is the vertical rise of the graph divided by the horizontal run.
$$\mathrm{slope} = \frac{\mathrm{rise}}{\mathrm{run}}$$ $$a = \frac{\Delta v}{\Delta t}$$ $$a = \frac{1}{4}$$hint
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.
solution
solution
Acceleration doesn't have to be constant. In the graphs below, acceleration is increasing. The change in acceleration over time is called jerk.
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.