Question 802783
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Let's say that at time *[tex \Large t_0] you are moving at *[tex \Large 4 \text{m/sec}] and your acceleration is *[tex \Large 1 \text{m/sec^2}].  Now, what is your velocity at *[tex \Large t_1] if *[tex \Large t\ =\ t_1\ -\ t_0\ =\ 1 \text{sec}]?  Well, *[tex \Large v_f\ =\ v_0\ +\ at], so putting in the numbers AND the units we get *[tex \Large v_f(\text{m/sec})\ =\ 4\text{m/sec}\ +\ 1\text{m/sec^2}*1\text{sec}].  Now, when you do the actual arithmetic, units in a numerator (as in the time) cancel units in a denominator (like one of the factors of 'sec' in acceleration).


Another way to look at it is that acceleration in meters per second squared times time in seconds has to equal velocity in meters per second.  Half of acceleration in meters per second squared times time in seconds squared is distance in meters.


It makes a lot more sense when you do it with the calculus (this for constant acceleration):


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  v(t)\ =\ \int\,a\,dt\ =\ at\ +\ v_o]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  s(t)\ =\ \int\,(at\ +\ v_o)dt\ =\ \frac{1}{2}at^2\ +\ v_ot\ +\ s_o]


And then back the other way:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ v(t)\ =\ \frac{ds}{dt}\ =\ at\ +\ v_o]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(t)\ =\ \frac{d^2s}{dt^2}\ =\ \frac{dv}{dt}\ =\ a]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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