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- anonymous

The position of an object at time t is given by s(t) = -4 - 2t. Find the instantaneous velocity at t = 6 by finding the derivative.

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- anonymous

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- anonymous

pleeeaaaaaase help

- amistre64

well, what is our derivative?

- anonymous

I have no idea how to start it that's the problem @amistre64

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- amistre64

your course materials should have a basic guideline, what have you covered?

- amistre64

how do you define a deriative?

- anonymous

I know but none of them look like this. They never went over a problem like this with velocity

- amistre64

lets start with how you define the derivative, what does your material say

- amistre64

does slope play into it?

- anonymous

yes

- amistre64

then give me your best understanding of the concept of a derivative

- anonymous

this is what we are suppose to use but it's confusing http://www.sosmath.com/calculus/diff/der00/der00.html

- amistre64

i agree that the technical part of it may be confusing, but the concept is pretty simple ... most concepts are.
the derivative can tell us the slope of a tangent line to a curve at any given point.
and it is said that a line is tangent to itself ...
does this mean anything to you?

- anonymous

Okay somewhat

- anonymous

If i followed the formula somewhat it would be -2 but that makes no sense to me

- amistre64

mathically, there is a long process that can be worked out with the limit of the difference quotient.
\[\lim_{h\to 0}\frac{f(x+h)-f(x)}{(x+h)-x}\]
let f(x) be a line ... mx+b
\[\lim_{h\to 0}\frac{m(x+h)+b-(mx+b)}{(x+h)-x}\]
\[\lim_{h\to 0}\frac{mx+mh+b-mx-b}{h}\]
\[\lim_{h\to 0}\frac{mh+b-b}{h}\]
\[\lim_{h\to 0}\frac{mh}{h}\]
\[\lim_{h\to 0}m\frac{h}{h}\]
\[\lim_{h\to 0}m\]
the derivative of a line, is just its slope, m

- anonymous

f(x)=−4−2x
f(x+h)=−4−2(x+h)
f′(x)=limh→0f(x+h)−f(x)h→limh→0−4−2(x+h)−(−4−2x)h

- amistre64

-2 is the derivative of your line equation. so yes. for any value of t, the derivative is a constant -2

- anonymous

ohh geez. I overcomplicated it

- amistre64

s(t) = -4-2t
v(t) = -2

- anonymous

Thank you!

- amistre64

good luck :)

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