Average acceleration and velocity problem

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Homework Help Overview

The problem involves determining the average acceleration of an object whose position is described by the function x = 4t^2 - 3t^3 over the interval from t = 0 to t = 2 seconds. Participants are exploring how to derive the average acceleration from the given position function.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to calculate instantaneous velocity using the derivative of the position function. There is confusion about how to apply the function to find the necessary values for velocity at specific times.

Discussion Status

Some participants have suggested calculating the derivative of the position function to find instantaneous velocities at t = 0 and t = 2. There is acknowledgment of the correct approach to find average acceleration, and one participant reports successfully obtaining the expected result after following the guidance provided.

Contextual Notes

Participants express uncertainty about using the given function to find velocity and average acceleration, indicating a need for clarification on the application of derivatives in this context.

newton9point8
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Homework Statement



Okay here's the problem, The coordinate of an object is given as a function of time by x = 4t^2 - 3t^3, where v is in m/s and t is in seconds. Its average acceleration over the interval form t = 0 to t = 2

I wast told the answer is -10m/s^2 but i can't figure out how to get the same anwer.

Homework Equations



Aavg = V2-V1/time

The Attempt at a Solution



ok I've already tryed to plug in 4*2^2 - 3*2^3 and 4*0^2 - 3*0^3 but it doesn't work...

i know if i can get the right values for the change of velocity over 2seconds (my time) i can get the right answer but i can't figure out how to obtain
the change of velocity with the given function :confused:
 
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Newton9point8 said:

Homework Statement



Okay here's the problem, The coordinate of an object is given as a function of time by x = 4t^2 - 3t^3, where v is in m/s and t is in seconds. Its average acceleration over the interval form t = 0 to t = 2

I wast told the answer is -10m/s^2 but i can't figure out how to get the same anwer.

Homework Equations



Aavg = V2-V1/time

The Attempt at a Solution



ok I've already tryed to plug in 4*2^2 - 3*2^3 and 4*0^2 - 3*0^3 but it doesn't work...

i know if i can get the right values for the change of velocity over 2seconds (my time) i can get the right answer but i can't figure out how to obtain
the change of velocity with the given function :confused:
Welcome to the forum.
First, let's start with a simpler question. Given x(t), do you know how to calculate th evelocity at a given time? (I am talking about the instantaneous velocity, not the average velocity). You need to know that because the first thing you need to do is to calculate the velocity at t=0 and then at t=2 seconds.
 
the instantaneus velocitiy formula is V= dx/dt but I still need to know what to do with the given function... i don't know how to use the given funtion to get the answer. If I was given the v1 and v2 in m/s i would just take the difference of v1 and v2 divide it by the time but that functions is what is holding me back. :(
 
Newton9point8 said:
the instantaneus velocitiy formula is V= dx/dt but I still need to know what to do with the given function... i don't know how to use the given funtion to get the answer. If I was given the v1 and v2 in m/s i would just take the difference of v1 and v2 divide it by the time but that functions is what is holding me back. :(

You just provided the answer yourself! To find the instantaneous velocity at a certain time, you need to calculate the derivative of th efunction x(t) and then evaluate it at that time. So to find v1 (= velocity at t=0). find dx/dt and then set t=0. To find v2, plug t=2 in the derivative dx/dt.
 
Got it!
so i got the derivative v=8t-9t^2 and then i got -20 after i pluged in my v1 and v2 then divided it by two and that gave me -10m/s^2! (right answer)

Thanks for your help! :biggrin:
 
Newton9point8 said:
Got it!
so i got the derivative v=8t-9t^2 and then i got -20 after i pluged in my v1 and v2 then divided it by two and that gave me -10m/s^2! (right answer)

Thanks for your help! :biggrin:

That's completely right. Good job.

Glad I could help.
 

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