I know how to figure out derivatives now, but I dont know what people say when they say 'second derivative of displacement'?
im trying to find velocity and acceleration...
also I dont know wtf "ds/dt" and stuff like that means?
05-24-2007, 02:43 AM #1
Math help: what is a derivative of displacement?
Last edited by whodat?; 05-24-2007 at 02:45 AM.
05-24-2007, 02:56 AM #2
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what ure looking for is :
displacement is basicly distance it can be represented as a function of s -
s = kt
where k is a variabe in terms of time t - we'll consider k as position.
then the rate of change of displacement = ds/dt will give us the rate of change of distance over time
or v(t) = velocity
if we look for the rate of change of velocity with repsect to time we get
dv(t) / dt = accelleration
so to recap
d/dt of displacement (or the rate of change of disp(or distance in a specified direction)) = velocity
d/dt of velocity (or the rate of change of velocity ) = acceleration
Last edited by iBswole; 05-24-2007 at 02:59 AM.
05-24-2007, 02:57 AM #3
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05-24-2007, 02:59 AM #4
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ahh i remember doing this cant remember what answer is though
ds/dt maybeStrength, size, power and speed arent achievable without dedication, focus and determination.
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05-24-2007, 03:02 AM #5
i know how to find the derivative, but how do i find the 'second deriv of displacement'???
s = -5t^2 + 120t + 500
i know the deriv is -10x+120
but i dont know what the second derivative of displacement is
Last edited by whodat?; 05-24-2007 at 03:08 AM.
05-24-2007, 03:12 AM #6
05-24-2007, 03:13 AM #7
05-24-2007, 03:23 AM #8
so she teaches it like we already know how to do it. plus she is indian so its hard to understand here
i know how to do pretty much everything except figure out acceleration, and acceleration is the second derivative of displacement and i dont know how to figure that out...
the problem is:
An object moves according to the distance (s), time (t) equation:
s = -5t^2 + 120t + 500
1) the average velocity between t=0 and t=3 seconds
answer: [delta]S / [delta]t which equals 105, got that problem right
2) the velocity equation (the derivative)
answer: -10x + 120
3) the velocity at t=2 seconds
answer: -10(2) + 120 = 100
4) the accelration equation
5) acceleration at t=3 seconds
6) time to reach max height
answer: havent figured out yet but I know how to
7) max height above ground
answer: havent figured out yet but i know how to
05-24-2007, 03:29 AM #9
05-24-2007, 03:31 AM #10
05-24-2007, 03:58 AM #11
05-24-2007, 04:01 AM #12
05-24-2007, 04:03 AM #13
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05-24-2007, 04:05 AM #14
I'm trying to follow the teacher and take notes and I'm just like 'W T F'
Then I see that example and its like... just double down? I was gettin all worked up when I just do it twice?
I learned more on friggin Yahoo! education in 10 minutes than I did in over 15 hours in that class..
Last edited by whodat?; 05-24-2007 at 04:09 AM.
05-24-2007, 04:15 AM #15
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go to www.mathhelpforum.com ... it really helped me through calc... people explain the answers and make everything very simple!!! You can thank me later
05-24-2007, 04:18 AM #16