Average Velocity between time to time

• Triple88a
In summary, the student attempted to find the average velocity between 2 and 4 seconds, but they incorrectly took their derivative. The website where they submitted their answers says the equation is wrong, but the average velocity is valid for constant acceleration only.
Triple88a

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.

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.

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?

Last edited:

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

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

Last edited:
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.

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).

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.

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. :)

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.

What is average velocity between time to time?

Average velocity between time to time is a measurement of the overall displacement of an object over a certain period of time. It takes into account both the direction and the magnitude of the object's motion.

How is average velocity between time to time calculated?

To calculate average velocity between time to time, you divide the total displacement of the object by the total time it took to make that displacement. The resulting value is the average velocity.

What is the difference between average velocity and instantaneous velocity?

Average velocity is a measure of the overall displacement of an object over a period of time, while instantaneous velocity is a measure of the object's velocity at a specific moment in time. Average velocity takes into account the entire motion, while instantaneous velocity only looks at a single point in time.

Can average velocity between time to time be negative?

Yes, average velocity between time to time can be negative. This means that the object has a net displacement in the opposite direction of its initial motion. It is important to note that negative velocity does not necessarily mean the object is moving backwards, as direction is also taken into account.

How does average velocity between time to time relate to average speed?

Average velocity and average speed are related, but they are not the same. Average speed is a measure of how fast an object is moving, while average velocity takes into account the object's direction of motion. For example, if an object travels 10 meters north and then 10 meters south in 5 seconds, its average speed would be 4 meters per second, but its average velocity would be 0 meters per second because its net displacement is 0.

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