# What is the definition of terminal speed

## 1. Path, time and speed

The simplest definition of speed is "distance per time", i.e. the quotient of the distance covered and the time required for it.

In physics, the path is usually shortened with the letter \ (s \) (note:

**S.**trecke) and the time with the letter \ (t \) (note:**t**ime). The speed is given the abbreviation \ (v \) (engl. "**v**elocity "). Then the connection is given byBecause \ (s \), \ (t \) and \ (v \) are not pure numbers, but physical quantities, the units used are important. Depending on the units in which the distance (i.e. a length) and the time are specified, the speed is given a different unit (which consists of a distance or length unit divided by a time unit).

Often larger distances are given in kilometers (km) and the time for the required journey in hours (h). Then you can calculate the speed from these two measurements directly in km / h (kilometers per hour - that actually means km divided by h), which is the most frequently used unit for speed in everyday life.

It is important that the units of distance, time and speed fit together; otherwise the formula above will not work for the conversion.

During its orbit around the sun, the earth covers about 30 km in space in one second, depending on where it is. If we calculate the speed from this information, we can on the one hand take over these units directly and obtain the speed from \ (s = 30 \, km \) and \ (t = 1 \, s \)

\ [v = \ frac {s} {t} = \ frac {30 \, \ rm km} {1 \, \ rm s} = 30 \ frac {\ rm km} {\ rm s} \]

in the unit km / s (kilometers per second). On the other hand, we might want to know how much that is in m / s or in km / h. In the first case, we need to convert the 30 km to meters and get it

\ [v = \ frac {30 \, \ rm km} {1 \, \ rm s} = \ frac {30 \, 000 \, \ rm m} {1 \, \ rm s} = 30 \, 000 \ , \ frac {\ rm m} {\ rm s} \,. \]

In the second case we can either assume that \ (1 \, {\ rm s} = \ frac {1} {3600} \, {\ rm h} \), or we consider that the earth is in a Hour travels 3600 times the distance of one second (since \ (1 \, {\ rm h} = 3600 \, {\ rm s} \)), and received

\ [v = 3600 \ cdot 30 \, \ frac {\ rm km} {\ rm h} = 108 \, 000 \, \ frac {\ rm km} {\ rm h} \ ,. \]

\ [v = \ frac {s} {t} = \ frac {30 \, \ rm km} {1 \, \ rm s} = 30 \ frac {\ rm km} {\ rm s} \]

in the unit km / s (kilometers per second). On the other hand, we might want to know how much that is in m / s or in km / h. In the first case, we need to convert the 30 km to meters and get it

\ [v = \ frac {30 \, \ rm km} {1 \, \ rm s} = \ frac {30 \, 000 \, \ rm m} {1 \, \ rm s} = 30 \, 000 \ , \ frac {\ rm m} {\ rm s} \,. \]

In the second case we can either assume that \ (1 \, {\ rm s} = \ frac {1} {3600} \, {\ rm h} \), or we consider that the earth is in a Hour travels 3600 times the distance of one second (since \ (1 \, {\ rm h} = 3600 \, {\ rm s} \)), and received

\ [v = 3600 \ cdot 30 \, \ frac {\ rm km} {\ rm h} = 108 \, 000 \, \ frac {\ rm km} {\ rm h} \ ,. \]

In the metric system, the basic units for length are

**the meter**and for the time the**Second (s)**. That is why the basic unit for speed in physics is a**Meters per second (m / s)**. To see how km / h converts to m / s, we express the kilometer and the hour as multiples of the base units:\ (1 km = 1000 m \) and \ (1 h = 3600 s \).

\ [1 \, \ frac {\ rm km} {\ rm h} = \ frac {1 \ rm \, km} {1 \ rm \, h} = \ frac {1000 \, \ rm m} {3600 \ , \ rm s} = \ frac {1} {3,6} \, \ frac {\ rm m} {\ rm s} \ ,, \]

where in the last step we reduced 1000 in the fraction.

So if we want to convert from km / h to m / s, we have to divide by 3.6. Conversely, we have to multiply by 3.6 when converting m / s to km / h, because if we bring 3.6 to the km / h side, we have

\ [1 \, \ frac {\ rm m} {\ rm s} = 3.6 \, \ frac {\ rm km} {\ rm h} \ ,. \]

If time and speed are not given, but time and speed, we can calculate the distance covered by solving and obtaining the formula at the top for \ (s \)

\ [s = vt \ ,. \]

Finally, we can also resolve this for the required time \ (t \) if the path \ (s \) and the speed \ (v \) are given:

\ [t = \ frac {s} {v} \ ,. \]

In these two equations, too, it is important to pay attention to the units used and, if necessary, to convert the values into more suitable units.

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