Decomposition of Distance and Time on Speed

Question

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$_1$ $= 1.2$

2. 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

Answer 1

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).

Answer 2

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.

Answer 3

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.