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UrbanXrisis
Sep20-04, 07:04 PM
This is an equation of a position v time graph:

x=342t^4-127t^3+1.87t^2+2.45t

I need to use this expression to find the instantaneous velocity at t=0.000s

arildno
Sep20-04, 07:06 PM
And how is inst. velocity related to position?

Tom McCurdy
Sep20-04, 07:08 PM
take the derivative of the first equation
then just plug in t
and volla you got instantious velocity

UrbanXrisis
Sep20-04, 07:17 PM
x`=1368t^3-381t^2+3.74t+2.45

is that correct?

Tom McCurdy
Sep20-04, 07:18 PM
This is what i got
its in white so you must highlight
1368x^3 - 381x^2+3.74X+2.45
=2.45=instant velocity

lol
oh t is 0
well this is how you would find instant acceleration at other times
4104x^2-762x+3.74

sorry I added extra but I am just learning this so I thought it would be good practice.

Tom McCurdy
Sep20-04, 07:19 PM
Oh sorry I guess tex equations don't turn white

arildno
Sep20-04, 07:21 PM
Besides, you used x as the variable rather than t :wink:

Tom McCurdy
Sep20-04, 07:25 PM
Darn I was so close to getting it right