Rate of change of acceleration

[LT72884]

Homework Statement

A particle traveled in a straight line in such a way that its distance (S) from a given point on that line after time (t) was S= 20t^3 -t^4 The rate of change of acceleration at time t=2 is what value?

Relevant Equations

S = 20t^3 - t^4

A particle traveled in a straight line in such a way that its distance (S) from a given point on that line after time (t) was S= 20t^3 -t^4 The rate of change of acceleration at time t=2 is what value?ok, I am kind of stuck on this very simple problem. It should be as simple as taking the derivative twice, and plugging in 2 at the end..
so if S = 20t^3 - t^4 then
V = 60t^2 - 4t^3 and
acceleration = 120t-12t^2 plug in 2 for t and it should be 240-48 = 192 but apparently the answer is 72... why?

am i not finding acceleration but really finding jerk? so i needed to do 3 derivs?
thanks

 

[ergospherical]

am i not finding acceleration but really finding jerk? so i needed to do 3 derivs?

Yep.

 

[LT72884]

Is it because it says find the rate of change of acceleration? If so, its one of them trick questions hahahaha

 

[ergospherical]

Defo: rate of change of acceleration $\equiv \dot{a} = \ddot{v} = \dddot{s}$

 

[LT72884]

Well, SOB, they done tricked me with their fancy words haha

thanks

 

[haruspex]

Is it because it says find the rate of change of acceleration? If so, its one of them trick questions hahahaha

It's not a trick, it's what rate of change means.
Given an expression for displacement as a function of time:
Instantaneous velocity = rate of change of displacement = 1st derivative of displacement
Instantaneous acceleration = rate of change of velocity = 2nd derivative of displacement
Instantaneous jerk = rate of change of acceleration = 3rd derivative of displacement
:
As far as you like.
You are asked for the rate of change of acceleration, not the instantaneous acceleration.

 

[sysprog]

Just chiming in on what @ergospherical and @haruspex have already said nicely: {assuming single vector here) first derivative of position wrt time is velocity; second is acceleration, third is jerk (rate of increase or decrease in acceleration), fourth is jounce (or snap, in case, no really, some guys enjoy this, you want to call the fifth and sixth crackle and pop -- as if anyone who doesn't listen to the musical explosions of his wilting rice krispies should care ) $-$ so yes, you really do have to figure the third derivative, and don't you of course step-by-step need the first and second to get to the third?

 

[PeroK]

but apparently the answer is 72...

$72 km/h$, $72 m/s$ or $72$ something else?

 

[sysprog]

$72 km/h$, $72 m/s$ or $72$ something else?

From: https://www.varsitytutors.com/calculus_1-help/how-to-find-acceleration

hmm $\dots$