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?


05242007, 02:43 AM #1
Math help: what is a derivative of displacement?
Last edited by whodat?; 05242007 at 02:45 AM.

05242007, 02:56 AM #2
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 ) = accelerationLast edited by iBswole; 05242007 at 02:59 AM.

05242007, 02:57 AM #3

05242007, 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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05242007, 03:02 AM #5
this makes no sense to me at all... i am math retarded or something
i know how to find the derivative, but how do i find the 'second deriv of displacement'???
example:
s = 5t^2 + 120t + 500
i know the deriv is 10x+120
but i dont know what the second derivative of displacement isLast edited by whodat?; 05242007 at 03:08 AM.

05242007, 03:12 AM #6

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05242007, 03:23 AM #8
its a summer school class, and all of the people in it except me are retaking it becaue they failed (im taking it to get ahead)
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
Find:
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
DONT KNOW
5) acceleration at t=3 seconds
DONT KNOW
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

05242007, 03:29 AM #9


05242007, 03:31 AM #10

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05242007, 04:01 AM #12

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05242007, 04:05 AM #14
hehehe.... moron explanation is what makes the most sense to me
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?; 05242007 at 04:09 AM.


05242007, 04:15 AM #15
math help...
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

05242007, 04:18 AM #16
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