Metric Units

Physical quantities can be expressed in terms of these metric base units:

distance: meter (m)
time: second (s)
mass: kilogram (kg)

We use metric prefixes to indicate multiplication or division by powers of ten. Converting from 10km to meters means moving the decimal 3 spaces larger.

$$10 \, \mathrm{km} = 10\left(10^{3}\right) \mathrm{m} = 10{,}000 \, \mathrm{m}$$ Converting from 10 ms to meters means moving the decimal 3 spaces smaller. $$10 \, \mathrm{ms} = 10\left(10^{-3}\right) \mathrm{s} = 0.01\, \mathrm{s}$$
Name Symbol Factor Power
giga G 1,000,000,000 10 9
mega M 1,000,000 10 6
kilo k 1,000 10 3
centi c 0.01 10 −2
milli m 0.001 10 −3
micro μ 0.000 001 10 −6
nano n 0.000 000 001 10 −9
Example: 10 km is how many meters?
solution $$\mathrm{k} = 1000$$ $$10\, \mathrm{km} = 10(1000)\, \mathrm{m} = 10{,}000\, \mathrm{m}$$
Example: 450 nm is how many meters?
solution $$\mathrm{n} = 10^{-9}$$ $$450\,\mathrm{nm} = 450\left( 10^{-9}\right) \mathrm{m} = 0.00000045 \, \mathrm{m}$$

Nonmetric Units

Converting outside the metric system is more complex. You can do a google search to find out how two units are related. You then build a conversion fraction with the new unit on top and the old unit on the bottom to be canceled out. The top and the bottom of the fraction must be equal to each other.

Lets convert 50 minutes into seconds.

$$50 \,\mathrm{min} \to \mathrm{s}$$ In order to cancel minutes we want to build a fraction with minutes on top and the seconds on the bottom. $$50 \,\mathrm{min} \left( \mathrm{\frac{s}{min}} \right)$$ The fraction must be equal to one to not change the value we are converting. This means the top and bottom must equal each other. One minute equals 60 seconds. $$50 \, \mathrm{min} \left( \frac{60\, \mathrm{s}}{1\,\mathrm{min}} \right)$$ We can cancel units. $$50 \, \textcolor{red}{ \mathrm{min}} \left( \frac{60\, \mathrm{s}}{1 \,\textcolor{red}{\mathrm{min}} } \right)$$ $$50 \left( \frac{60\, \mathrm{s}}{1} \right)$$ Multiply and divide to simplify $$3000\, \mathrm{s}$$
Example: A marathon is 26.2 miles. If one mile is the same as 1.6 km, how far is a marathon in kilometers?
solution $$26.2 \, \textcolor{red}{ \mathrm{mile}} \left( \frac{1.6 \, \mathrm{km}}{1 \,\textcolor{red}{ \mathrm{mile}}} \right) = 41.92\, \mathrm{km}$$
Example: I'm 6 feet and 1 inch tall. How many meters tall am I?
solution $$\mathrm{ft \to inches}$$ $$6\, \mathrm{ft}\left( \frac{12\,\mathrm{in}}{1\, \mathrm{ft}}\right) + 1\, \mathrm{in}$$ $$72\, \mathrm{in} + 1\, \mathrm{in}$$ $$73\, \mathrm{in}$$
$$\text {a google search returns: } 1\mathrm{m} = 39.37 \mathrm{in}$$ $$73\,\textcolor{red}{ \mathrm{in}} \left( \frac{1\, \mathrm{m}}{39.37 \, \textcolor{red}{ \mathrm{in}}}\right)$$ $$73 \left( \frac{1\, \mathrm{m}}{39.37}\right)$$ $$1.85\, \mathrm{m}$$

Position

In physics we use the symbol x for position. That means you shouldn't use x for an unknown unless you are solving for position. We generally use meters(m) as our units.

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

$$\Delta x = x_{f}-x_{i}$$

$$\Delta x$$ = change in position, distance, displacement, length [m, meters]

$$x_i$$ = initial (starting) position [m, meters]

$$x_f$$ = final (ending) position [m, meters]

Time

We keep track of time in the same way as position. We use t for time and generally use seconds(s) as our units.

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

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

$$t_i$$ = initial (starting) time [s, seconds]

$$t_f$$ = final (ending) time [s, seconds]

Velocity

Velocity is a measure of how much position changes (Δx) over a period of time (Δt).

$$v = \frac{\Delta x}{\Delta t}$$

$$\Delta x$$ = distance [m, meters]

$$\Delta t$$ = time period [s, seconds]

$$v$$ = average velocity [m/s, meters per second]

Example: A car is traveling at 20 m/s for 80 seconds. How far does the car travel?
solution \begin{aligned} v &= \small\frac{\Delta x}{\Delta t} \\ 20\, \mathrm{\tfrac{m}{s} }&= \small\frac{\Delta x}{80\, \mathrm{s}} \\ (20 \mathrm{\tfrac{m}{s}} )(80\, \mathrm{s}) &= \Delta x \\ 1600\, \mathrm{m} &= \Delta x \end{aligned}
Example: Someone tells you they can run a 10 km race in about an hour. What velocity is that in m/s? Is that fast? (1 m/s is walking speed)
solution $$10 \, \mathrm{km} = 10(1,000) \mathrm{m} = 10,000 \, \mathrm{m}$$ $$1\, \mathrm{h} \left(\frac{60\, \mathrm{min}}{1\, \mathrm{h}}\right)\left(\frac{60\, \mathrm{s}}{1\, \mathrm{min}}\right) = 3,600 \mathrm{s}$$ $$v = \frac{\Delta x}{\Delta t}$$ $$v = \frac{10,000\, \mathrm{m}}{3,600\, \mathrm{s}}$$ $$v = 2.7\, \mathrm{\tfrac{m}{s}}$$
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 on average? Answer this one in miles/hour.
solution $$v = \frac{\Delta x}{\Delta t}$$ $$v = \frac{270\, \mathrm{miles} }{4\, \mathrm{hour}}$$ $$v = 67.5\, \mathrm{\tfrac{miles}{hour} }$$
Example: If I walk at a speed of 1.2 m/s how long will it take for me to walk 2 km?
solution $$2 \, \mathrm{km} = 2(1000) \mathrm{m} = 2,000 \, \mathrm{m}$$ $$v = \frac{\Delta x}{\Delta t}$$ $$\Delta t = \frac{\Delta x}{v}$$ $$\Delta t = \frac{2,000\, \mathrm{m}}{1.2\, \mathrm{\tfrac{m}{s}}}$$ $$\Delta t = \frac{2,000}{1.2}\, \mathrm{s}$$ $$\Delta t = 1,666.\overline{6} \, \mathrm{s}$$

Acceleration

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

$$a = \frac{\Delta v}{\Delta t}$$

$$\Delta v$$ = change in velocity [m/s] = $$v_f-v_i$$

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

$$a$$ = acceleration [m/s²]

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 car can go from 0 to 60 miles per hour in 5.9 s. What is the acceleration in m/s²? (1 mile = 1609 meters)
solution $$60\left( \mathrm{ \frac{\color{red}{mile}}{\color{blue}{hour}}} \right)\left(\frac{1609\,\mathrm{ m}}{1 \,\color{red}{\mathrm{mile} }}\right)\left(\frac{1\, \color{blue}{ \mathrm{hour} }}{3600\, \mathrm{s} }\right) = 26.8 \mathrm{\tfrac{m}{s}}$$ $$a = \frac{\Delta v}{\Delta t}$$ $$a = \frac{26.8\,\mathrm{ \tfrac{m}{s}}} {5.9\, \mathrm{s}}$$ $$a = 4.5\, \mathrm{ \tfrac{m}{s^{2}} }$$
Example: I start a velocity of 1 m/s. I speed up to 3 m/s over 10 seconds. What is my acceleration?
solution $$\Delta v = v_{f}-v_{i}$$ $$\Delta v = 3 \,\mathrm{ \tfrac{m}{s}} -1\, \mathrm{ \tfrac{m}{s} }$$ $$\Delta v = 2\, \mathrm{\tfrac{m}{s}}$$
$$a = \frac{\Delta v}{\Delta t}$$ $$a = \frac{2\, \mathrm{\tfrac{m}{s}} }{10\, \mathrm{s} }$$ $$a = 0.2\,\mathrm{ \tfrac{m}{s^{2}}}$$
Example: These graphs show velocity vs. time. What do their slopes represent?
solution

acceleration

acceleration is defined as the change in velocity every second. This is the same as the slope of the graphs.

Using Proportions

If you change a variable we can predict whether other variables in the same equation will increase or decrease. The first step is to decide if the relationship between the variables is directly proportional or inversely proportional.

$${\color{#09e}{\Uparrow} \atop y} {\atop = k } {\color{#09e}{\Uparrow} \atop x} \quad\quad\quad {\color{#f00}{\Downarrow} \atop y} {\atop = k } {\color{#f00}{\Downarrow} \atop x} \quad\quad\quad \frac{ y \color{#f00}{\Downarrow}}{x \color{#f00}{\Downarrow}} = k$$

directly proportional = a constant ratio, k, between two variables. Both variables increase and decrease together.

$${\color{#09e}{\Uparrow} \atop y} {\color{#f00}{\Downarrow} \atop x} {\atop = k } \quad\quad\quad {\color{#f00}{\Downarrow} \atop y} {\color{#09e}{\Uparrow} \atop x} {\atop = k } \quad\quad\quad y {\color{#f00}{\Downarrow}} = \frac{k}{x \color{#09e}{\Uparrow}}$$

inversely proportional = a constant ratio, k, between a variable and the inverse of a variable. Variables increase and decrease opposite to each other.

You can use proportions when everything else in an equation is constant except two variables. Then you predict how one will change when the other changes.

$$v = \frac{\Delta x}{\Delta t}$$

Example: You can run 100 m in about 15 s. If you want to decrease your time, should you increase or decrease your velocity?
solution $$\text{Δx is constant}$$ $$v = \frac{\Delta x}{\Delta t}$$ $$v \Delta t = \Delta x$$ $$\text{v and Δt are inversely proportional}$$ $${\color{#09e}{\Uparrow} \atop v} {\color{#f00}{\Downarrow} \atop \Delta t} {\atop =} { \atop \Delta x}$$ $$\text{increasing v will decrease Δt}$$

$$F = ma$$

Example: A force of 100 N produces an acceleration of 10 m/s² for a 10 kg mass. How should you change the force to reduce the acceleration?
solution $$\text{m is constant}$$ $$F = ma$$ $$\text{F and a are directly proportional}$$ $${\color{#f00}{\Downarrow} \atop F} {\atop =} { \atop m} {\color{#f00}{\Downarrow} \atop a}$$ $$\text{Decreasing F will decrease a}$$