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I integrated a jerk-function three times (acceleration, velocity, position) to get the resulting function $s(t)$ for the position. I am not sure how to use the SI units or dimension in the function.

This is my function. Putting in the time $t$ as a simple number the result will be the meters the object has passed.

$$\ s(t) = 2t³ - 14t² + 20t - 44 .$$

Is this correct:

$$\ s(t) = 2t³ [m/s³] - 14t² [m/s²] + 20t [m/s] - 44 [m].$$

Sorry if this is the wrong site for this kind of question as it is more physics related. I also feel like this is way to obvious but I can't come up with a solution that seems right to me.

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  • $\begingroup$ "Putting in the time $t$ as a simple number", but then the attempt assumes that $t$ is not unitless but with "seconds" unit. $\endgroup$
    – peterwhy
    Commented Jan 20 at 2:39
  • $\begingroup$ It's weird to mix things like acceleration, jerk, velocity and distance like that. I think it should be $(2t^3-14t^2+20t-44)[m]$. You worked up from $m/s^3$ to $m$ by integrating. $\endgroup$
    – CyclotomicField
    Commented Jan 20 at 2:43
  • $\begingroup$ As far as I know, $s(t) = 2t^3 [m/s^3] - 14t^2 [m/s^2] + 20t [m/s] - 44 [m]$ is correct. However, in my opinion, nobody really cares if you simply write $s(t) = 2t^3- 14t^2+ 20t- 44$, provided that you put the correct units when a specific value of $t$ is set. For example, if $t=1s$, then we substitute $t=1$ into $s(t) = 2t^3- 14t^2+ 20t- 44$ to get $s(1)=-36$, but in the end you must write in the correct units $\boxed{s(t=1s)=-36m}$. $\endgroup$
    – QuantumSuperfield
    Commented Jan 20 at 3:18
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    $\begingroup$ Step by step, if you are given $j(t) = 12 \text{ m/s$^3$}$ (a constant), then integrating once gives $a(t) = \left(12 \text{ m/s$^3$}\right)t + C_1$, where $C_1 = -28 \text{ m/s$^2$}$ is to be determined (e.g. if the initial acceleration is given). At each step the coefficients keep their dimensions. $\endgroup$
    – peterwhy
    Commented Jan 20 at 3:27

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$t$ and $s(t)$ are physical quantitities. Quoted from Wikipedia:

A physical quantity can be expressed as a value, which is the algebraic multiplication of a numerical value and a unit of measurement.

Your equation $$s(t) = 2t^3 - 14t^2 + 20t - 44$$ is not correct to begin with. $s(t)$ is a length (having unit $\text{m}$) and $t$ is a time (having unit $\text{s}$). Hence the equation above is dimensionally inhomogeneous.

You make it dimensionally homogeneous by writing $$s(t) = 2\frac{\text{m}}{\text{s}^3}\ t^3 - 14\frac{\text{m}}{\text{s}^2}\ t^2 + 20\frac{\text{m}}{\text{s}}\ t - 44\text{ m}$$

When you now insert a value for $t$ with unit $\text{s}$ (for example $t=4.2\text{ s}$), you see the $\text{s}^3$, $\text{s}^2$ and $\text{s}$ in the terms on the right side all cancel out, and you finally get a number with unit $\text{m}$ which matches the unit of $s(t)$ on the left side.

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