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# Thread: Math help: what is a derivative of displacement?

1. ## Math help: what is a derivative of displacement?

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?

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 ) = acceleration

3. Originally Posted by whodat?
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?

If you have a function let say y=x which is a func for displacement

First Derivate will be y'=1 (is a function for VELCOITY)
Second Derivative will be y''=0 (is a function for ACCELERATION)
Third Derivative will be y'''= 0(for argument sake)(is a function for JERK)

4. ahh i remember doing this cant remember what answer is though

ds/dt maybe

5. Originally Posted by iBswole
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
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 is

6. Is that physics?

7. Originally Posted by MesoPeaks
Is that physics?
no, math class...practical application problem

8. Originally Posted by MesoPeaks

Can you post some of your lecture..?
its a summer school class, and all of the people in it except me are re-taking 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)

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

9. I found a website that explained it.. correct me if I'm wrong...

but all I do is take the origional derivative and break it down again?

**** if its that ez then i was going ape over nothing

10. what website..?

11. Originally Posted by MesoPeaks
what website..?
http://www.nipissingu.ca/calculus/tu...er_higher.html

12. y'- means first derivative
y"-means second derivative
y'"-third
y""-72x their right, so 72<---------fourth derivative

13. yes, for the second derivative, just take the derivative of the derivative

so -10t+120 -> -10t

14. Originally Posted by bigbondo
yes, for the second derivative, just take the derivative of the derivative

so -10t+120 -> -10t
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..

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

16. Originally Posted by xcelxp
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
thanks for the link, i'll keep that bookmarked

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