Convert the acceleration $10m{s^{ - 2}}$ into $km{h^{ - 2}}$

Hint: As, we know that $1km = 1000m$, then $1m = \dfrac{1}{{1000}}km$ and we also know that $1h = 60\min = 60 \times 60\sec = 3600\sec $, then $1\sec = \dfrac{1}{{3600}}h$, on substituting the values we will get the required result.

Complete step-by-step answer:
We, have to convert the acceleration $10m{s^{ - 2}}$ into $km{h^{ - 2}}$
For this, we know that
$1km = 1000m$
And $1h = 60\min = 60 \times 60\sec = 3600\sec $
But as we have to convert the $m$ into $km$ so we can write it as, $1m = \dfrac{1}{{1000}}km$
And seconds into hours so we can write, $1\sec = \dfrac{1}{{3600}}h$
Now, as acceleration is $a = 10m{s^{ - 2}}$
Now convert meter into kilometer and seconds into hours by using the above relations, we get
$ \Rightarrow a = 10\dfrac{m}{{{s^2}}}$
$ \Rightarrow a = \dfrac{{10 \times \dfrac{1}{{1000}}}}{{{{\left( {\dfrac{1}{{3600}}} \right)}^2}}}$
$ \Rightarrow a = \dfrac{{10 \times {{\left( {3600} \right)}^2}}}{{1000}}$
$ \Rightarrow a = 36000 \times 36$
$ \Rightarrow a = 1296000$
$ \Rightarrow a = 1.296 \times {10^6}km{h^{ - 2}}$
This is the required acceleration in $km{h^{ - 2}}$

Additional information:
Definition of kilometer: a kilometer is a decimal multiple of the meter, the international unit of the length, approximately equal to 39370.07 inches. A kilometer is now used officially for expressing distance between two places or points.
Definition of meter: A meter is the base unit of the length in the international system of units. Therefore, 1km=1000 meters.
Definition of hours: A period of time equal to 60 minutes that is 3600 seconds.
Definition of seconds: it is measured as the smallest unit of time. Therefore, $1\sec = \dfrac{1}{{3600}}h$

Note: for this problem, there is a standard equation for this type of conversion. To convert a kilometer per hour into meters per second, we need to multiply by 5 and divide by 18. This standard form got from this $\dfrac{{\dfrac{{1km}}{{1000}}}}{{\dfrac{{1{h^2}}}{{3600 \times 3600}}}}$ fraction which we used in the solution to solve. When we solve this fraction, we will get above term.