# Derivative as a rate of change

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• Oct 25th 2009, 11:12 PM
VitaX
Derivative as a rate of change
$s = -t^3 +3t^2 -3t$ with the interval $0\leq t \leq 3$

a. Find the body's displacement and average velocity for the given time interval.

$V=-3t^2 +6t - 3$
$V(0) = - 3$
$V(3) = -32$
Is this all thats needed on this part or do I need to do $V(3)-V(0)$ to find the average velocity?

b. Find the body's speed and acceleration at the endpoints of the interval.

$s = -t^3 +3t^2 - 3t$
$s(0)= 0$
$s(3)=9$ is positive because speed is always absolute value
$A = -6t + 6$
$A(0) = 6$
$A(3) = -12$

c. When, if ever, during the interval does the body change direction.

Use the Velocity equation for this I believe and when I plug in values between the interval, the output (v) is never changed into a positive number so can I conclude it does not change direction?
• Oct 25th 2009, 11:23 PM
Arturo_026
for a) , average velocity means distance/time so find [s(3)-s(0)]/(3-0). And displacement will just be distance, i.e. s(3) - s(0)

for b) they ask you for absolute value of v(t) at the endpoints. and also a(t)

Don't forget that: s'(t) = v(t) ; s''(t) = a(t) = v'(t).
• Oct 25th 2009, 11:32 PM
VitaX
Quote:

Originally Posted by Arturo_026
for a) , average velocity means distance/time so find [s(3)-s(0)]/(3-0). And displacement will just be distance, i.e. s(3) - s(0)

for b) they ask you for absolute value of v(t) at the endpoints. and also a(t)

Don't forget that: s'(t) = v(t) ; s''(t) = a(t) = v'(t).

For a) wouldn't it be v as the variable and not s? But for b) im cofused at why you say they ask for the absolute value of v(t) because didn't I solve that in a) and doesn't it ask to just find the speed and acceleration at the end points?
• Oct 25th 2009, 11:37 PM
Arturo_026
It seems that you are confusing what the s, v, and t stand for. Here:
s(t), the function you were given is the position function, it denots where the point is with respect to time t.
v(t) is s'(t) and that's the velocity of the point at that instant t. And when they ask you for speed, it simply means the absolute value of v(t).
and a(t) denotes acceleration.
• Oct 25th 2009, 11:40 PM
VitaX
Quote:

Originally Posted by Arturo_026
It seems that you are confusing what the s, v, and t stand for. Here:
s(t), the function you were given is the position function, it denots where the point is with respect to time t.
v(t) is s'(t) and that's the velocity of the point at that instant t. And when they ask you for speed, it simply means the absolute value of v(t).
and a(t) denotes acceleration.

Could you show me your answers for a, b, and c if its not too much trouble so I can fully understand what you are saying
• Oct 25th 2009, 11:41 PM
Arturo_026
Quote:

Originally Posted by VitaX
For a) wouldn't it be v as the variable and not s?

v is the velocity and s is the position, so to find the average velocity, as I said is distance/time , think of it as driving from point a to b, and they are 20 miles apart, then it takes you 2 hours to get there, therefore your average speed was 10 miles per hour.
• Oct 26th 2009, 12:23 AM
VitaX
Quote:

Originally Posted by Arturo_026
for a) , average velocity means distance/time so find [s(3)-s(0)]/(3-0). And displacement will just be distance, i.e. s(3) - s(0)

for b) they ask you for absolute value of v(t) at the endpoints. and also a(t)

Don't forget that: s'(t) = v(t) ; s''(t) = a(t) = v'(t).

I understand what you are saying now about |v(t)| = speed i overlooked that part in the book and i was thinking that speed was just obtained by plugging in values into the original equation. But my final question is when you write to find the displacement and average velocity is that just plugging the endpoint values of the interval into the given equation or plugging them into v(t) (the derivative of the given equation)
• Oct 26th 2009, 05:10 AM
HallsofIvy
The average velocity between t=0 and t= 3 is (s(3)- s(0))/(3- 0), the distanced moved between those two times, divided by the change in time.
• Oct 26th 2009, 04:59 PM
VitaX
Quote:

Originally Posted by HallsofIvy
The average velocity between t=0 and t= 3 is (s(3)- s(0))/(3- 0), the distanced moved between those two times, divided by the change in time.

Ok then I guess I just use the given equation for part a. then
• Oct 26th 2009, 05:25 PM
Open that Hampster!
If it helps make it semantically easier to understand, the derivative/rate of change of displacement is velocity, and the derivative/rate of change of velocity is acceleration. That makes sense if you think about it.

(A) Average displacement uses the original equation/3.

Average velocity uses its derivative, but you miscalculated V(3). It should be -12.

(B) Looks correct

(C) Your assumption is correct. The body will change direction if the sign of the velocity function changes within the interval.
• Oct 26th 2009, 07:54 PM
VitaX
Quote:

Originally Posted by Open that Hampster!
If it helps make it semantically easier to understand, the derivative/rate of change of displacement is velocity, and the derivative/rate of change of velocity is acceleration. That makes sense if you think about it.

(A) Average displacement uses the original equation/3.

Average velocity uses its derivative, but you miscalculated V(3). It should be -12.

(B) Looks correct

(C) Your assumption is correct. The body will change direction if the sign of the velocity function changes within the interval.

Isn't displacment just $s(3) - s(0)$ and then average velocity $\frac{v(3)-v(0)}{3-0}$ If this is correct then all is cleared up :D
• Oct 26th 2009, 08:11 PM
Arturo_026
Quote:

Originally Posted by VitaX
Isn't displacment just $s(3) - s(0)$ and then average velocity $\frac{v(3)-v(0)}{3-0}$ If this is correct then all is cleared up :D

I understand that you are having a hard time understanding it, but try reading the chapter in your book that talks about that.

You are correct about displacement, but average velocity is displacement/time ; that is (s(0)-s(3))/(0-3) ; I understand if it doesn't make sense to you since it says average velocity and you have distance in your numerator, but trust me that's what it is.
• Oct 26th 2009, 08:13 PM
VitaX
Quote:

Originally Posted by Arturo_026
I understand that you are having a hard time understanding it, but try reading the chapter in your book that talks about that.

You are correct about displacement, but average velocity is displacement/time ; that is (s(0)-s(3))/(0-3) ; I understand if it doesn't make sense to you since it says average velocity and you have distance in your numerator, but trust me that's what it is.

Ok I trust you. So average velocity uses the given equation. but when it says find the velocity at the end points of the interval means find the derivative of the given equation and go from there. If thats correct than everything is cleared up. Just some terminology has me confused. mainly part a.
• Oct 26th 2009, 08:14 PM
Arturo_026
Quote:

Originally Posted by VitaX
Ok I trust you. So average velocity uses the given equation. but when it says find the velocity at the end points of the interval means find the derivative of the given equation and go from there. If thats correct than everything is cleared up. Just some terminology has me confused. mainly part a.

Yes.
• Oct 26th 2009, 08:16 PM
Open that Hampster!
Apologies for my earlier post; it's been awhile since I've taken Physics.

The unit for velocity is in $\frac{m}{s}$

Which is distance over time. Average velocity is just the average of the distance over the average of time.

Though in this example you're lucky, because V(3)-V(0) = S(3) - S(0)
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