# More on velocity and speed

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The s=v/t formula

We have that

$\bar v = \frac{{\Delta s}}{{\Delta t}}$

For linear motion with constant velocity and if we assume the starting position is at 0 we may write this as

$v = \frac{s}{t}$

We can essentially hear this definition everytime we say a speed. The car is driving at 100 km/h, i.e. 100 km per hour, 100 km every hour,

$v = \frac{{100\;{\rm{km}}}}{{1\;{\rm{h}}}} = 100\;{\rm{km/h}}$

More examples:

If an object moves 100 m in 20 s then the speed is 100/20 = 5.0 m/s.

If we travel 1800 km in 2.0 h then our speed is 1900/2.0=900 km/h, so we are most probably flying.

Velocity

We use the same formula, but with the displacement instead of travelled distance. We also have to consider the direction.  Say we define north to be in the positive direction, and that we travel straight south 1800 km in 2 h, then we have travelled at an  average velocity of  –900 km/h.

Then a bit of algebra

Suppose we instead know the time taken and the distance and want to find the time? We then need to make t the subject in v=s/t. If we want to do this step by step then we start by multiplying both sides by t to get vt=s or

s=vt

As we will see in a moment this is a useful formula too. Next we divide by v to get

$t = \frac{s}{v}$

Let us say we travel 1800 km at 900 km/h, how long did it take? We get 1800/900=2.0 h.  Oh, it is the previous example, but now in another order.

Let us return to s=vt. Suppose we travel at 900 km/h for 2.0 h, how far do we get? We get 900·2=1800 km.

A look at units

Let us look a bit more carefully on the last two examples, but now with the units in the equations too.

$\begin{array}{*{20}{c}} {t = \frac{s}{v} = \frac{{1800\;{\text{km}}}}{{900\;{\text{km/h}}}} = \frac{{1800\;{\text{km}} \cdot {\text{h}}}}{{900\;\frac{{{\text{km}}}}{{\text{h}}} \cdot {\text{h}}}} = \frac{{1800\;{\text{km}} \cdot {\text{h}}}}{{900\;\frac{{{\text{km}}}}{{\left. {\underline {\, {\text{h}} \,}}\! \right| }} \cdot \left. {\underline {\, {\text{h}} \,}}\! \right| }}} \\ {\quad \quad = \frac{{1800\;{\text{km}} \cdot {\text{h}}}}{{900\;{\text{km}}}} = \frac{{1800\;\left. {\underline {\, {{\text{km}}} \,}}\! \right| \cdot {\text{h}}}}{{900\;\left. {\underline {\, {{\text{km}}} \,}}\! \right| }} = \frac{{1800\;}}{{900\;}}{\text{h}} = 2.0\;{\text{h}}} \end{array}$

(The parts indicated by $\left. {\underline {\, {\text{ }} \,}}\! \right|$  cancel against each other). One normally don’t do algebra on units as detailed as this, but we do it here once to see that it actually work out right in the end.

You might have seen a “helping” svt-triangle that is supposed to help you figure hut how the three formulas

$v = \frac{s}{t},\quad s = vt,\quad t = \frac{s}{v}$

could be found. It might be helpful initially, but honestly, it will be an hindrance to learning in the long run. If you need the triangle (that I won’t write here) to change one of the formulas to another one of them, then you really ned to practice on your elementary algebra.

To use the svt-triangle is as to learn tricks on how to play the piano with your elbows without knowing to play with your fingers. It might work for very simple tunes, but it will definitely be a problem when trying to play Chopin’s  Nocturnes.

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