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For calculation of displacement based on acceleration and time, I have the formula below:

$$ x = v_0t + \tfrac{1}{2}at^2 .$$

If I get $a$ in km/h$^2$, and $x$ in km and $v_0$ in km/h and I suppose to calculate the $t$ in hours, will the formula above be valid? or the term $\tfrac{1}{2}$ will have to change?

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3 Answers 3

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Yes, if you use same units then there will be no change. If you use all measurements in fps then also formula remains valid. It is universal, given all units are in same scale.

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    $\begingroup$ Please use proper spelling. Using "u" when you mean "you" detracts from an otherwise correct answer. $\endgroup$
    – Floris
    Commented Jun 29, 2015 at 16:43
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    $\begingroup$ Sorry didn't noticed that one..edited $\endgroup$
    – Aneek
    Commented Jun 29, 2015 at 16:54
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First of all, as already mentioned, yes $t$ is in units of hours. An equility remains an equality after introducing units. Hence, the unit on the RHS and the LHS have to agree and in fact also the sum of two quantities (as $3.6m/s+ 1km/h$) can only be simplified when both have the same unit!

Clarifications of the appearance of units
Units are actually a sign of physics, while the formula above is a simple mathematical solution of a differential equation (insert $x(t)=v_0 t+ \tfrac{1}{2}a t^2$ and you find it fulfills this equation, try it without the $1/2$) $$ \frac{d^2 x(t)}{d t^2}=a $$ Giving physical meaning to the quantities in your equation $x(t)=v_0 t+ \tfrac{1}{2}a t^2$ is then the process of performing natural science, i.e. identifying $x(t)$ with the position (of a car let's say) at an instance of time $t$, whereby at $t=0$ one measures a velocity of $v_0\equiv \frac{dx(t)}{dt}\big|_{t=0}$ and an accelaration of $a\equiv \frac{d^2x(t)}{dt^2}\big|_{t=0}$.

Experiments are almost exclusively of relative nature, that means you measure (so one measures) a quantity w.r.t. some scale:

  • The position is measured w.r.t. a length scale $\mu_{x}$ (e.g. $\mu_x\equiv 1m$ as defined by the SI system)
  • Time is measured w.r.t. a time scale $\mu_{t}$ (e.g. $\mu_t\equiv 1s$ as defined by the SI system), etc.

Imagine every day observations without a reference scale: I drove 20 today, take 10 flour, or so, which is not graspable at all.

Now the physical quantities are measured in multiplicities of these scales, resulting in: $\tilde{x}(t)\equiv \mu_x x(t)$, $\tilde{t}\equiv \mu_t t$ $$x(t)=v_0 t+ \tfrac{1}{2}a t^2 \iff \tilde{x}(t)/\mu_x=v_0 \tilde{t}/ \mu_t+ \tfrac{1}{2}a (\tilde{t}/ \mu_t)^2 $$ Mutliplying by $\mu_x$ (e.g. $\mu_x=1m$) yields $$ \tilde{x}(t)=(v_0 \cdot \frac{\mu_x}{\mu_t})\tilde{t}+ \tfrac{1}{2}(a \cdot \frac{\mu_x}{\mu_t^2} )\tilde{t}^2 $$ And thus finally $\tilde{v}_0\equiv v_0 \cdot \frac{\mu_x}{\mu_t}$ and $\tilde{a}\equiv a \cdot \frac{\mu_x}{\mu_t^2} $ are the physical velocity and acceleration in units of $\mu_x$ and $\mu_t$: $$ \tilde{x}(t)=\tilde{v}_0\tilde{t}+ \tfrac{1}{2}\tilde{a}\tilde{t}^2 $$

The $\frac{1}{2}$ is due to a mathematical property, while the units are introduced by some reference experiment (measurement) as for instance the experiments you find in the definition of the SI system. Once you choose for $v_0=1km/h$ and $a=1m/s^2$ you have to convert one of them into the same system of units to define $\mu_t$ and $\mu_x$ and thus the unit of the quantities $x(t)$ and $t$.

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As long as your units are consistent you can use any units you want and the equation will be unchanged. A change would only be required if you started mixing up the units e.g. specifying the velocity in m/s and the acceleration in km/h$^2$.

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  • $\begingroup$ So does it mean the if I solve for $t$, it will be in hours $\endgroup$
    – Dumbo
    Commented Jun 29, 2015 at 16:22
  • $\begingroup$ @Sean87 Yes it will. You have to keep all the units consistent for the math to work. $\endgroup$
    – Floris
    Commented Jun 29, 2015 at 16:39

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