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} |
solution
$$ \mathrm{k} = 1000 $$ $$10\, \mathrm{km} = 10(1000)\, \mathrm{m} = 10{,}000\, \mathrm{m}$$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 $$3000\, \mathrm{s}$$solution
$$26.2 \, \textcolor{red}{ \mathrm{mile}} \left( \frac{1.6 \, \mathrm{km}}{1 \,\textcolor{red}{ \mathrm{mile}}} \right) = 41.92\, \mathrm{km}$$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]
solution
$$\begin{aligned} v &= \small\frac{\Delta x}{\Delta t} \\ 20\, \mathrm{\tfrac{m}{s} }&= \small\frac{\Delta x}{80\, \mathrm{s}} \\ (20 \mathrm{\tfrac{m}{s}} )(80\, \mathrm{s}) &= \Delta x \\ 1600\, \mathrm{m} &= \Delta x \end{aligned}$$solution
$$10 \, \mathrm{km} = 10(1,000) \mathrm{m} = 10,000 \, \mathrm{m}$$ $$1\, \mathrm{h} \left(\frac{60\, \mathrm{min}}{1\, \mathrm{h}}\right)\left(\frac{60\, \mathrm{s}}{1\, \mathrm{min}}\right) = 3,600 \mathrm{s}$$ $$v = \frac{\Delta x}{\Delta t}$$ $$v = \frac{10,000\, \mathrm{m}}{3,600\, \mathrm{s}}$$ $$v = 2.7\, \mathrm{\tfrac{m}{s}}$$solution
$$v = \frac{\Delta x}{\Delta t}$$ $$v = \frac{270\, \mathrm{miles} }{4\, \mathrm{hour}}$$ $$v = 67.5\, \mathrm{\tfrac{miles}{hour} }$$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]
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
$$ 60\left( \mathrm{ \frac{\color{red}{mile}}{\color{blue}{hour}}} \right)\left(\frac{1609\,\mathrm{ m}}{1 \,\color{red}{\mathrm{mile} }}\right)\left(\frac{1\, \color{blue}{ \mathrm{hour} }}{3600\, \mathrm{s} }\right) = 26.8 \mathrm{\tfrac{m}{s}} $$ $$a = \frac{\Delta v}{\Delta t}$$ $$a = \frac{26.8\,\mathrm{ \tfrac{m}{s}}} {5.9\, \mathrm{s}}$$ $$a = 4.5\, \mathrm{ \tfrac{m}{s^{2}} }$$solution
$$\Delta v = v_{f}-v_{i}$$ $$\Delta v = 3 \,\mathrm{ \tfrac{m}{s}} -1\, \mathrm{ \tfrac{m}{s} }$$ $$\Delta v = 2\, \mathrm{\tfrac{m}{s}}$$$$a = \frac{\Delta v}{\Delta t}$$ $$a = \frac{2\, \mathrm{\tfrac{m}{s}} }{10\, \mathrm{s} }$$ $$a = 0.2\,\mathrm{ \tfrac{m}{s^{2}}}$$
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.
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.