Average Velocity between time to time

AI Thread Summary
To find the average velocity between time 2 and time 4 for the position equation P = -5t^3 + 5t^2 - 4t, the correct method involves calculating the change in position over the change in time, rather than using the derivative. The values for position at t=2 and t=4 are -28 and -256, respectively, leading to a total change in position of -228 over a time interval of 2 seconds. This results in an average velocity of -114. The initial calculation of -124 was incorrect due to misunderstanding the average velocity formula, which should reflect the total change in position divided by the total time elapsed. Understanding this concept is crucial for solving similar physics problems.
Triple88a
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Homework Statement


I have to find the average velocity of an equation from time 2 to time 4. This equation here is of the position.
P = -5*t^3+5*t^2 -4*t.
2. The attempt at a solution

Since this is the position i have to take derivative of that.
-15t^2-10t-4

Then i substitute 2 for t and i get -44 then i substitute 4 and get -204

( -44 + -204) / 2 = -124 average velocity. Is this correct? The website where i submit my answers to says it's wrong.
Thanks in advance.
 
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Your equation for average velocity is valid for constant acceleration only. In general, average velocity must be calculated using \Delta s/\Delta t
 
Triple88a said:

Homework Statement


I have to find the average velocity of an equation from time 2 to time 4. This equation here is of the position.
P = -5t^3+5t^2 -4t

2. The attempt at a solution

Since this is the position i have to take derivative of that.
-15t^2-10t-4

Then i substitute 2 for t and i get -44 then i substitute 4 and get -204

( -44 + -204) / 2 = -124 average velocity. Is this correct? The website where i submit my answers to says it's wrong.
Thanks in advance.

It looks like you took your derivative wrong. I think the second term should be positive, ie -15t^2+10t-4
 
Bacat said:
It looks like you took your derivative wrong. I think the second term should be positive, ie -15t^2+10t-4


i'm sorry that was a typo
but yes though the position formula is = 5*t2 -4*t -5*t3.

thats 10t-4-15t^2
10x2 - 4 - 15x2^2
20 - 4 - 15x4
16 - 60
-44

and

10x4 - 4 - 15x4^2
40 - 4 - 240
36 - 240
-204

(-44 + -204)/2
-124

so what is the average velocity between 2 and 4 seconds?
and i punch in -124 and nothing :(
 
i found it by guessing starting at -100 and going negative one by one.. it said ok at -113

how can that be?
 
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Try re-reading my first response.:wink:
 
If you have access to calculus, just use mean value theorem.
If not, listen to phantom jay.
 
Jay is right. Since acceleration is changing in time you just need to divide the total change in position by the total time, ie

\frac{\Delta S}{\Delta T} = \frac{-256-(-28)}{2} = -113
 
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Bacat said:
Jay is right. Since acceleration is changing in time you just need to divide the total change in position by the total time, ie

\frac{\Delta S}{\Delta T} = \frac{-256-(-28)}{2} = -113

hhmm so what goes where? I'm just starting physics. Its actually my 4th day in the class and we have like 3 formulas that are all for constant acceleration. The funny thing is what is on our homework is much more advanced vs what we're learning in class.
 
  • #10
Maybe the letters are confusing you.

In physics, \Delta S means Total change in position. And \Delta T means Total time elapsed. So to get the -256 number you just plug in 2 for t in your position equation (the one you were given). To get the -28 number, you plug in 4 for t.

Physically, these two numbers are your position in coordinate space at t=2 and t=4. You want the difference between these two values (as indicated by \Delta) so you subtract them. You find the total time elapsed by subtracting the values you plugged in for t (4 - 2 = 2).

Think about it this way:

If you take 1 hour to walk 5 miles, you walked an average of 5 miles per hour. It doesn't matter how many times you stopped or started...whether you ran really fast sometimes and then slower other times...overall you took 1 hour to go 5 miles. So by dividing 5 miles by 1 hour you get the average velocity of 5 miles per hour.

You're doing the same thing here except that instead of hours you're using seconds. Instead of "5 miles" you are given an equation which plots your position. But it's the same concept exactly.

Cheers.
 
  • #11
Ahh that makes sense. i get it now :)
Thank you all for your responses. I appreciate the help a ton.

This forum is definitely going in the favorites. :)
 
  • #12
Triple88a said:
hhmm so what goes where? I'm just starting physics. Its actually my 4th day in the class and we have like 3 formulas that are all for constant acceleration. The funny thing is what is on our homework is much more advanced vs what we're learning in class.
Well, I see that things haven't changed since I last took Calc based Physics 201 some 40 years ago. The teaching concentration by far is on constant acceleration problems, but then they throw the other harder stuff at you , and you're on your own. But sometimes, it's the best way to learn.
 
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