Average acceleration and velocity problem

AI Thread Summary
The problem involves finding the average acceleration of an object with the position function x = 4t^2 - 3t^3 over the interval from t = 0 to t = 2 seconds. To determine average acceleration, the instantaneous velocities at t = 0 and t = 2 must be calculated by taking the derivative of the position function, resulting in v = 8t - 9t^2. After evaluating this derivative at the specified times, the change in velocity is found to be -20 m/s. Dividing this change by the time interval of 2 seconds yields the average acceleration of -10 m/s^2, confirming the correct answer. The discussion emphasizes the importance of using derivatives to find instantaneous velocity for calculating average acceleration.
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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