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I would like to know the individual effect of changing distance and time on the change of speed.

Eg. 1. Distance $= 1200$

Time $= 1000$

Speed$_1$ $= 1.2$

  1. Distance $= 1400$

    Time $= 1200$

    Speed$_2$ = $1.16667$

I understand that multiplying by proportional difference in distance and time will give me the new speed.

Speed$_1 $ (1.166667) / (1.2) = $ $Speed$_2$

I'm having issues getting figures so that:

Speed$_1$ $+/-$ effect of distance $+/-$ effect of time = Speed$_2$

eg. $1.2 + 0.2 - 0.233333 = 1.16667$

Could someone please explain how to do this and the math to support it?

Sorry if this is a bit too basic for this forum or is poorly worded.

Thanks

ett
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SlinkyBill
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  • It's impossible to do in a simple way. If you want to have the effect from time and the effect from distance completely separated into two independent changes, then you have seen yourself that what you use those changes for is multiplying and dividing. Addition and subtraction is different from multiplying and dividing on a fundamental level, so you can't just change which one you want to use. – Arthur Nov 24 '17 at 16:05

3 Answers3

0

This is very basic mathematics.

You have speed $s$, distance $d$ and time $t$.

$s=\frac dt$

If $d$ becomes $d_2=a*d$, then $s_2=\frac{d_2}{t}=\frac{a*d}{t}=a*s$

If $t$ becomes $t_3=b*t$, then $s_3=\frac{d}{t_3}=\frac{d}{b*t}=\frac sb$

You can combine the two effects.

Martigan
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Using Martigans variables, it appears you are hoping to write
$s_2=s+(a-1)s$ and call $(a-1)s$ the effect of distance, then write $s_3=s+s(\frac 1b-1)$ and call $s(\frac 1b-1)$ the effect of time. This does not work in that if we multiply the distance by $a$ and the time by $b$ the new speed is $s_4=\frac {as}b \neq s+(a-1)s+s(\frac 1b-1)$

You can do what you are asking as long as $b$ is very close to $1$. Then $\frac 1b-1\approx 1-b$ and $s_4\approx s+(a-1)s+(1-b)s$ is in the form you desire.

Ross Millikan
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Okay, it seems that you know the formula $$s=\frac{d}{t}\tag{$\star$}$$ for average speed $s$ over a distance $d$ and time $t.$ It also seems that, given speed, distance, and time $s_1,d_1,t_1$ (respectively) such that $s_1=\frac{d_1}{t_1},$ and given some arbitrary distance $d$ and arbitrary non-zero time $t,$ you're looking for functions $f$ and $g$ such that if $$s:=s_1+f(d,d_1)+g(t,t_1),\tag{$\heartsuit$}$$ then $(\star)$ holds. (I'm less sure about this part, as your post is a bit unclear.)

Clearly, if $d=0,$ then we need to have $s=0,$ regardless of $t,$ or $(\star)$ will fail to hold. It then follows from $(\heartsuit)$ that for $(\star)$ to hold, we need $$g(t,t_1)=-s_1-f(0,d_1)\tag{1}$$ for all $t.$ For $(\star)$ to hold whenever $(\heartsuit)$ does, then by $(\heartsuit)$ and $(1),$ we need $$s=f(d,d_1)-f(0,d_1)\tag{2}$$ for all $d.$ Another thing we need in order for $(\star)$ to hold is that, when $d=d_1,$ then $s=\frac{t_1}{t}s_1,$ regardless of $t.$ By $(2),$ we then need $$\frac{t_1}{t}s_1=f(d_1,d_1)-f(0,d_1)\tag{!}$$ for all $t,$ but this is impossible unless $s_1=0,$ for otherwise the left-hand side will vary as $t$ does, while the right-hand side remains constant.

Therefore, since we derived a contradiction from our assumption that $f$ and $g$ existed such that $(\star)$ holds whenever $(\heartsuit)$ holds, then no such functions $f$ and $g$ exist.

Upshot: The reason you're having trouble getting the figures you're looking for is that it's impossible (assuming I understand correctly what you're trying to do).

Cameron Buie
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  • Thanks for the answers Cameron and Ross! They clearly explain why what I am trying to do isn't possible.

    I am an analyst and was trying to apply the concept of s = d/t to the equation ROI = revenue/cost, which is why my physics ability is limited.

    From your explanation it's clear on why my figures didn't work out when using s4=asb≠s+(a−1)s+s(1b−1).

     – SlinkyBill Nov 27 '17 at 09:20