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We know $\vec F =m\vec a$

Then $m\vec a =\frac {d\vec v}{dt}$

If this equality holds, this means direction of velocity vector and acceleration vector is same.

How this could be possible? What mistake I have done?

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    $\begingroup$ The mistake you have done is that the direction of $\bf{v}$ and its derivative can be in the opposite. For example v is parabolic, its derivative can decrease while it increases. $\endgroup$ – Turgon Feb 2 '17 at 6:21
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If this equality holds ,this means direction of velocity vector 
and acceleration vector is same .

No, it does not.

First of all, there is a mistake in your formula. Correct one is $\vec a =\frac {d\vec v}{dt}$. (Note, there is no $m$ in here). And this is just a definition of acceleration. This formula is NOT derived from $\vec F =m\vec a$. As I said it's just a definition of acceleration.

There is no reason why $\frac {d\vec v}{dt}$ and $\vec v$ should have the same direction. We have a body, it's velocity is $\vec v$ at the moment. A little bit later it's velocity is a little different: $\vec v + \vec v_1$. The difference $\vec v_1$ can be directed anywhere, isn't it? For simplicity let's assume the acceleration is constant. In this case the difference is proportional to time: $\vec a * dt$. So, the direction of $\vec a$ is the direction of difference of velocity. It can have any direction!

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You have misunderstood the equation: $$ \vec{a} =\frac{d\vec{v}}{dt} $$ In fact, it means that the acceleration has the same direction of the variation of the velocity, and not of the velocity itself.

As a simple, unidimensional example, think of a particle moving along the $x$ direction and decelerating. Its velocity will be directed as $\hat{u}_x$ (unitary vector directed along the $x$ axis), but its acceleration will be directed as $-\hat{u}_x$.

Another simple example is given by an object in circular uniform motion. As you can see from this image when the object moves of an angle $θ$, its velocity changes from $v_1$ to $v_2$. Analyzing these two vectors, you can write the variation of this velocity, the $Δv$ vector, that will have neither vectors' direction. In fact, it will be directed towards the center, where we know should be a force (as a rope, or an electromagnetic interaction...) and therefore an acceleration that keep the system in this motion.

This answer would like to be direct and intuitive, not mathematically rigorous. A real proof of what I have said would come from analyzing how the operation of deriving acts on the direction of a vector.

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