In the equation: $a = dv/dt$ , is $dt$ the time taken to achieve that instantaneous acceleration? If you solve for $dt$ from $a = \frac{dv}{dt}$ , is it the time taken to to achieved that instantaneous acceleration?
$a$ : acceleration
$v$ : velocity
$t$ : time 
 A: No, it is not.
Suppose, a body is moving at a uniform velocity $v$, now there is no restriction on how much time it wants to remain with that same velocity. And after sometime it can accelerate if there is a net force on it. Now, acceleration means a rate of change in velocity and obviously it will take some time to increase (or decrease) it's velocity. It is therefore given as $$a=\frac{\Delta v}{\Delta t}$$  This acceleration may not be uniform and in principle can change. Therefore, you would want to know what exactly was acceleration at a particular time. This happens when you take the limit of $\Delta t$ tending to zero, so that time duration over which change of velocity happens is as short as possible to get the most accurate result which gives the instantaneous acceleration. This differential small time is denoted by $dt$.
I hope this makes things clearer.
A: In the equation $a= \frac{dv}{dt}$, $dt$ is actually the time in which that small change in the velocity of the body is brought.
So you can say that it is the time to achieve that very acceleration.
But actually acceleration is defined as the change in velocity in a certain amount of time. Mathematically you can get time if acceleration is given but you can not define acceleration without knowing about time. So meaningfully $dt$ is not the time to achieve that acceleration but it is the time in which that acceleration is achieved. It can sound confusing but think of it for once.
Thanks for asking.
