0
$\begingroup$

If we have a function, which describes, how a displacement in space along a line varies as a function of time: e.g.: $s(t)=vt$, its units are meters because $[v]=\frac{\text{meters}}{\text{seconds}}$ and $[t]=\text{seconds}$ thus $[vt]=\frac{\text{meters}}{\text{seconds}}\text{seconds}=\text{meters}$

A linearly accelerated motion would have the units $\mathrm{\frac {\frac ms} s=\frac m{s^2}}$

A variably accelerated motion (jerk) could have the units $\mathrm{\frac {\frac {\frac ms} s}s=\frac m{s^3}}$

What units would the variable acceleration described by the function $v*sin(\frac t{t_{MAX}})$ have ?

EDIT:
The meaning of that last function is supposed to be "a speed along a straight line, whose magnitude varies from $0$ to $v$ (as time varies from 0 to $t_{MAX}$), with the $v(t)$ "shape" like the sine function which varies from 0 to 1". Consider only the domain from $0$ to $\frac\pi 2$.

$\endgroup$
  • 2
    $\begingroup$ Trigonometric functions take angles (dimensionless) as arguments (at least I have not seen anything different). So, you would have to multiply by something with inverse units of t (e.g. angular frequency, $\omega$, $sin(\omega t)$). $\endgroup$ – Vangi Aug 7 '19 at 14:05
  • $\begingroup$ Do you actually want to say "the variable acceleration $resulting\ from\ a\ speed$ described by the function v⋅sin(t)" ? $\endgroup$ – EigenDavid Aug 7 '19 at 14:18
  • $\begingroup$ @David: Yes. In other words: a speed along a line, whose magnitude varies as the sine of time. $\endgroup$ – George Robinson Aug 7 '19 at 15:04
  • $\begingroup$ @Vangi: I don't think the argument, that a sine function takes, must represent an angle. If sine bother you, substitute a logarithm for it ...it would not change the nature of that question $\endgroup$ – George Robinson Aug 7 '19 at 15:07
  • $\begingroup$ @GeorgeRobinson But then you just derive twice your expression, not forgetting to write a pulsation ($\sin(\omega t)$); and you end up with the unit of the jerk, $ms^{-3}$ $\endgroup$ – EigenDavid Aug 7 '19 at 15:09
2
$\begingroup$

With the last revision, you have the expression $v \sin (t/t_{MAX})$ where I assume that $t_{MAX}$ is a time in consistent units. (Given the rest of the question, those units should be seconds.)

That has the same units as $v$, which in your case is a velocity with units of m/s.

The corresponding acceleration is $$ \frac{d}{dt} \left[ v \sin (t/t_{MAX}) \right] = \frac{v}{t_{MAX}} \cos(t/t_{MAX}) $$ and the corresponding jerk (time-derivative of acceleration) is $$ \frac{d^2}{dt^2} \left[ v \sin (t/t_{MAX}) \right] = -\frac{v}{t_{MAX}^2} \sin(t/t_{MAX}) .$$

Note the importance of having the dimensions of the argument of the trig functions correct, i.e. dimensionless. Those factors of $t_{MAX}$ in the derivative make the dimensions of acceleration and jerk work out to m/s$^2$ and m/s$^3$, respectively, which would not have happened in your original formulation where the argument was just "$t$".

If I define $\omega = 1/t_{MAX}$, then you could adopt units where $\omega = 1$. But you need to do it consistently such that you are no longer measuring time in seconds but rather in $\omega s$ (where $\omega$ has units of 1/s and the "s" in the last expression is the second). If you do that consistently, it's ok to write $\sin t$ in those units but then your velocity will, in consistent units, be in meters not meters-per-second.

$\endgroup$
1
$\begingroup$

The function $v\sin t$ as a rate of change of acceleration must have the unit $\rm m\, s^{-3}$ if it is to be dimensional correct.

A way of doing this is to write the function as $k \, v\sin t$ where the constant $k$ is equal to $1\,\rm s^{-2}$.

$\endgroup$
0
$\begingroup$

A variable acceleration could not be described by $v\times \sin(t)$ because you don't have the right units. A variable acceleration is $da/dt$. So it needs to have units of acceleration/time.

$\endgroup$
  • $\begingroup$ I think in this case the value of $sin(t)$ is just a dimensionless number that varies the velocity $v$ with time, when multiplied by it, so we clearly have a velocity varying in time. How would you correct it to make it dimensionally correct? $\endgroup$ – George Robinson Aug 7 '19 at 15:13
  • $\begingroup$ Part of the problem is that $\sin t$ is nonsense unless you've got units where time is dimensionless. If you have those units, then $\sin t$ is dimensionless, but that will also change the units of velocity $v$ in some compatible way. Otherwise, you need something like $\sin \omega t$ where $\omega$ also has units. You've put the question in an inherently inconsistent framework. @GeorgeRobinson $\endgroup$ – Brick Aug 7 '19 at 15:33
  • $\begingroup$ I know my framework is disorganized. Please help me improve it. I just intend to vary the magnitude of the speed in the same manner as the sine function varies. How to write this intention correctly? Would $v*sin(\frac t{t_{MAX}})$ make the argument to the sine function a dimensionless ratio? $\endgroup$ – George Robinson Aug 7 '19 at 15:46
0
$\begingroup$

Maybe it's worthwile to spell out in more detail what has been said in a comment by Steeven above:

Your post seems a bit confused (or is written in a confusing manner) about how units work. For example, a 'motion' doesn't have units.

Actually, it is basically simple:

  1. Displacement (distance travelled) has units of metres, $\text{m}$.

  2. Velocity is the change of distance per time and hence has units of distance per time, i.e. metre per second, $\text{m}/\text{s}$.

  3. Acceleration is the change of velocity per time, hence it has units of velocity per time, i.e. metre per second per second, $\text{m}/\text{s}^2$.

  4. Jerk is the change of acceleration per time, ... , units $\text{m}/\text{s}^3$.

You can see the pattern here. (Note that you can of course express distance in metres, miles, furlongs or whatnot, and time in seconds, hours or fortnights, and substitute any of these units for metres and seconds above. What is important is that you start with a unit of length, and each time derivative adds a unit of time in the denominator. Let us stick with metres and seconds for the rest of this answer.)

So acceleration always has units of $\text{m}/\text{s}^2$, regardless of whether it is constant or not. What distinguishes a motion with constant acceleration from on with variable acceleration is that the acceleration is, well, consatnt, and that thus the jerk, which measures the rate of change of the acceleration, is zero.

$\endgroup$
-3
$\begingroup$

Before downvoting please note that the OP in its original form stated that there was an expression giving the rate of change of accelleration as $v sin(t)$ and it is perfectly valid for this expression to give the quantum of the rate of change of accelleration.

Edited answer:

All expressions of a rate of change of accelleration over time need to have m per second per second per second units $m/s^3$.

If you can write the expression that describes it then it must have those units.

We can do various calculations on this to derive other values, such as the current accelleration:

We know the rate of change of the current accelleration is given by $v sin(t)$ so the current accelleration (units $m/s^2$) is given by the integral of this: $-v cos(t) +a0$ where $a0$ is the initial accelleration. Units are in $m/s^2$ as we integrated an expression in $m/s^3$ with respect to time.

So the current velocity (units $m/s$)is given by the integral of this: $a0t -vsin(t) + v0$ where $v0$ is the initial velocity. Units are in $m/s$ for the same reason as above.

The current displacement (units $m$)) is given by: $a0t^2/2 + v0t + vcost(t) + x0$ where $x0$ is the initial linear displacement. Units are in $m$ for the same reason as above.

My previous answer resulted in a lot of comment below, where I had made a mistake and not answered the OP question directly. I did not address the issue that the OP had already described the variable accelleration function.

The application of the mathematical functions sine and cosine here do not change the units, but the calculus (integration with respect to time) does.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.