Instantaneous rate of change

1. Sep 6, 2007

fitz_calc

1. The problem statement, all variables and given/known data

s is the position of a particle along the s-axis, find an expression for the velocity and acceleration and determine when v=0

s=1-3t^2

2. Relevant equations

v=ds/dt , a=dv/dt

3. The attempt at a solution

No idea where to begin, my book is not very clear. Do I just set v=0 equal to s = 1-3t^2? I know the answer is v=-6t but do not know where to begin - thanks!

2. Sep 6, 2007

Staff: Mentor

Yes, differentiate s(t) as you have shown to find v(t). Then set v(t) = 0, and solve for t. You're on the right track.

3. Sep 6, 2007

fitz_calc

I still do not know where to begin. my book gives an example s=2t^2-4t. when they begin to solve the problem they go from v=ds/dt to v=4t-4

i don't see how they come to this value?

4. Sep 6, 2007

PowerIso

They found the derivative of 2t^2 -4t.

5. Sep 6, 2007

fitz_calc

so i found acceleration to be -6 by taking the derivative of -6t. what does this mean in layman's terms, exactly? I don't understand how taking the derivative of s yields the velocity, and taking the derivative of velocity yields acceleration...

6. Sep 6, 2007

Staff: Mentor

Velocity is the change in position per unit time, or ds/dt. Acceleration is the change in velocity per unit time, or dv/dt.

** units of position s are meters [m]

** units of velocity v are meters per second [m/s]

** units of acceleration a are meters per second squared [m/s^2]

When you get an answer like a = -6 [m/s^2], that means that you have a *deceleration* of 6 [m/s^2]. Does that help?

Last edited: Sep 6, 2007
7. Sep 6, 2007

HallsofIvy

Staff Emeritus
It means that this object has a constant downward acceleration, just a something falling in constant gravity does.

Surely you have learned that the derivative is the rate of change of a function relative to the variable? If x(t) is distance, x, as a function of time, t, then dx/dt is the rate of change of distance relative to time: precisely the instantaneous velocity. Of course, the second derivative then is the rate of change of velocity relative to time: the definition of acceleration.

8. Sep 6, 2007

fitz_calc

i get it now, some of these topics were covered in my mechanics course last year, though we never really touched on derivatives like this technical calculus course is. thanks.

9. Sep 7, 2007

HallsofIvy

Staff Emeritus
Oh, God, I hate courses like that. I once taught a "Calculus for Economics and Business Administration" course. The text book I was required to use covered limits in one page just listing the three limit laws:
If $\lim_{x\rightarrow a} f(x)= L_1$ and $\lim_{x\rightarrow a} g(x)= L_2$ then $\lim_{x\rightarrow a} f(x)+ g(x)= L_1+ L_2$.
If $\lim_{x\rightarrow a} f(x)= L_1$ and $\lim_{x\rightarrow a} g(x)= L_2$ then $\lim_{x\rightarrow a} f(x)g(x)= L_1L_2$
If $\lim_{x\rightarrow a} f(x)= L_1$ and $\lim_{x\rightarrow a} g(x)= L_2$ AND $L_1$ is not 0, then $\lim_{x\rightarrow a} f(x)/g(x)= L_1/L_2$.

On the very next page, they defined the derivative as
$$\frac{df}{dx}(a)= \lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}$$
not bothering to point out that the laws of limits they had given do not apply here because the denominator necessarily does go to 0!

They had left out the most important law of limits:
If f(x)= g(x) for all x except x=a, then $\lim_{x\rightarrow a} f(x)= \lim_{x\rightarrow a}g(x)$!