Where is the element of time in the definition of work? In $\text{Work} = m \times a \times \text{distance}$.
From this equation, How can we know the duration of the force that have been applied
in order to move the object over that distance?
 A: Work is defined as $W=\int \vec F\cdot d\vec \ell$.  It may be that the force is  time-dependent so let's write $\vec F=\vec F(t)$.  You can then introduce time explicitly by writing 
$$
d\vec \ell =\frac{d\vec \ell}{dt}dt = \vec v(t) dt\, .
$$
In this case you simply end up with 
$$
W=\int \vec F(t)\cdot \vec v(t) dt\, .\tag{1}
$$
This last expression can be useful in some cases to show that a force does no work when it is perpendicular to $\vec v(t)$.  Thus the magnetic force $\vec F_{mag}= q\vec v\times \vec B$ is automatically perpendicular to $\vec v$ by the properties of the cross product, and can be seen to do no work as
$$
\left(q\vec v\times \vec B\right)\cdot \vec v \equiv 0\, .
$$
A: In the simplest case of a straight line movement with a constant force starting from a zero speed, from energy conservation
$$W=\dfrac{mv^2}{2}=\dfrac{m(2v_a)^2}{2}=2m\left(\dfrac{d}{t}\right)^2$$
Where $v_a$ is the average speed. The solution for time follows
$$t=\sqrt{\dfrac{2m}{W}}\cdot d$$
A: First of all: in general you cannot substitute $F_r=ma$ into $W=Fd$, the reason being that the work equation can be applied for any force, while Newton's 2nd law refers to the resultant force $F_r$. Also, if the force $F$ isn't constant, and we are in 3D, then work is calculated as $W=\int \vec{F}\cdot d\vec{x}$.
As for where's time in $W=Fd$, there's the time it takes for the distance $d$ to be covered, but, yes, time isn't really playing a role in this equation and, in general:


*

*you cannot know the duration of the interaction from the work of the force alone.


Time does appear in the related concept of power (time rate of work: $P_\text{avg}=W/\Delta t$). So you can calculate $\Delta t$, given total $W$ and $P_\text{avg}$.
