1. Simplification and Conversion 2. Application of functions in formulas

Simplifying and Converting  for transition of units a conversion unit is needed  the conversion factor allows units to be cancelled out through simplification  the end product is in the left over units

Simplifying and Converting (Continued)  The reason this works is because the conversion factor is equal to one.  For example:  Km m  3km x 1000m 1 1km = 3000m

Conversion Example 2  When dealing with derived units more than one conversion factor is required.  For example:  mi/h f/s  35 mi x 1h x 1min x 5280ft 1h 60min 60sec 1mi = 51.3 f/s

Functions  A function’s purpose is to substitute a numerical value for a variable.  For example:  F(6) = x = 8 F(6) = 8

Formulas  In physics the equivalent of a function is a formula.  Substitution value of desired variable  v = x/t (velocity = distance/time) x = 20ft t = 2s  v = 20ft/2s  v = 10ft/s

Formulas (Continued)  Linear programming – 2 variables  f(x,y) represents the limits (coordinates)  substitution of coordinates into the linear function will produce the desired value

Application in Physics  In physics sometimes two variables need to be replaced to find the answer.  For example:  f = ma (force = mass x acceleration)  m = kg  a = 24.5 m/s/s  15,192kg x 24.5m/s/s  thrust of rocket = 372,204N = 372,204N

Fantastic Failure  Miscalculation leads to disaster and unhappiness… Click here to watch video

Sweet Success  &feature=related &feature=related With your newly obtained prowess in math you can successfully launch yourself into outer space. Click here to watch video