can someone explain what s is in the following limit?
height of said brush at t seconds
The "s(t)" stands for a position function. Does this help?
at \(t=2,s(t)=-16\times 2^2+100=36\) and so you want \[\frac{36-(-16t^2+100)}{2-t}\]
we get \[\frac{36+16t^2-100}{2-t}=\frac{16t^2-64}{2-t}\] \[=\frac{16(t^2-4)}{2-t}=\frac{16(t+2)(t-2)}{2-t}\] \[=-16(t+2)\] replace \(t\) by 2 to get your answer
oh i get the s part, but if i put 2 in for t, that would give me zero...
you get \(\frac{0}{0}\) until you factor and cancel
okay got it! thanks! :)
once you factor and cancel you can replace \(t\) by 2
btw it always works this way, although in one week you will be taught a snap method and will be able to do this in your head on 3 seconds, and then get annoyed that you had to do it this way
you will say "the derivative of \(-16t^2+100\) is \(-32t\) replace \(t\) by \(2\) get \(-64\)
haha i think derivatives are for calc, im only in precal
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