anonymous
  • 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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
pleeeaaaaaase help
amistre64
  • amistre64
well, what is our derivative?
anonymous
  • anonymous
I have no idea how to start it that's the problem @amistre64

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amistre64
  • amistre64
your course materials should have a basic guideline, what have you covered?
amistre64
  • amistre64
how do you define a deriative?
anonymous
  • anonymous
I know but none of them look like this. They never went over a problem like this with velocity
amistre64
  • amistre64
lets start with how you define the derivative, what does your material say
amistre64
  • amistre64
does slope play into it?
anonymous
  • anonymous
yes
amistre64
  • amistre64
then give me your best understanding of the concept of a derivative
anonymous
  • anonymous
this is what we are suppose to use but it's confusing http://www.sosmath.com/calculus/diff/der00/der00.html
amistre64
  • 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
  • anonymous
Okay somewhat
anonymous
  • anonymous
If i followed the formula somewhat it would be -2 but that makes no sense to me
amistre64
  • 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
  • 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
  • amistre64
-2 is the derivative of your line equation. so yes. for any value of t, the derivative is a constant -2
anonymous
  • anonymous
ohh geez. I overcomplicated it
amistre64
  • amistre64
s(t) = -4-2t v(t) = -2
anonymous
  • anonymous
Thank you!
amistre64
  • amistre64
good luck :)

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