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If $F=0$, $F=ma \implies m\frac{(v-u)}{t}=0\implies v-u=0 \implies v=u $

Which is essentially Newton's first law.

The Cryptic Cat
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ACuriousMind's link gives you the correct answers.

But here is an additional side-note to be considered:

$$a= \frac{v-u}{t}\;;$$

Hmm; this is not true in general.

What you would want to say is that

$$F~=~ m\frac{\mathrm dv}{\mathrm dt} = ma~=~0$$

However, that doesn't mean that the body has constant velocity; for in some cases

$$\dot a ~\ne~ 0\;.$$

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