Calculating instantaneous acceleration

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SUMMARY

The discussion focuses on calculating the instantaneous acceleration of a particle moving along the x-axis, described by the position function x = 9.75 + 1.50t³. To find the instantaneous acceleration at t = 2.00 seconds, participants emphasize the necessity of taking the derivative of the position function twice. The first derivative yields the velocity function, while the second derivative provides the acceleration. A participant initially miscalculated the instantaneous velocity as 0.217 m/s, highlighting the importance of correctly applying calculus principles in physics problems.

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maxalador
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Homework Statement



The position of a particle moving along the x-axis is given in centimeters by x = 9.75 + 1.50t3, where t is in seconds

Calculate the instantaneous acceleration at 2.00 s.

Homework Equations


x = 9.75 + 1.50t3

I don't understand how i would solve for the acceleration using an equation for distance
i understand Vf=Vo+at and that the instantaneous velocity at this time is .217 m/s i had to teach myself how to do this so i am not sure i found the velocity correctly
 
Last edited:
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The way this works in PF is that you are supposed to make an attempt to solve the problem first, and show us what you did so far. Then, if necessary, we can try to get you pointed in the right direction.

Chet
 
Chestermiller said:
The way this works in PF is that you are supposed to make an attempt to solve the problem first, and show us what you did so far. Then, if necessary, we can try to get you pointed in the right direction.

Chet

Thanks for the info
 
You need to derive it. By t subject, do it twice. One derive tell you velocity, second tell you acceleration.
 
How did you arrive at v(2.00s) = 0.217 m/s? Are you familiar with what the derivative of position means?
 
Bandarigoda said:
You need to derive it. By t subject, do it twice. One derive tell you velocity, second tell you acceleration.

how would i derive it the second time
 
maxalador said:
how would i derive it the second time

You have a function for position with respect to time. Take the first time derivative. Show us what you get and what does this new function represent?
 
Clever-Name said:
How did you arrive at v(2.00s) = 0.217 m/s? Are you familiar with what the derivative of position means?

Did u ask it from me? I can't understand, sorry my bad english. Am i wrong about the derivation?
 
Bandarigoda said:
Did u ask it from me? I can't understand, sorry my bad english. Am i wrong about the derivation?

No, sorry, that was directed at maxalador.
 
  • #10
Clever-Name said:
No, sorry, that was directed at maxalador.

Oh okay. :)
 
  • #11
sorry guys it turns out that the reason i didnt understand was because i was doing it wrong. thanks for all your help though
 

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