By equation of motion :
[tex]2as=v^2-u^2\\\\s=\dfrac{v^2-u^2}{2a}\\\\s=\dfrac{0^2-2^2}{2\times( -0.2)}\\\\s=10\ m[/tex]
Now, it is given that it stops after 5 seconds of motion and time at that point is 12:00.
So, time when it started is 11 : 59 : 55.
Therefore, distance travelled is 10 m.
Hence, this is the required solution.
