# 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

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.

• Thanks. When we say dv, dx etc. etc. , does the ‘d’ stand for taking the derivative or does ‘d’ just signify a very very small quantity? Or is ds just s2- s1 on a displacement time graph but the displacement is so small that we denote it by using ‘d’ as ds? Or is it just the derivative? I’m new to calculus so sorry for the seemingly easy question which I do not seem to get. Commented May 23, 2020 at 14:35
• @Curious $d$ before any quantity means that you want an infinitesimal change ( infinitely small change ) in some quantity like velocity, distance or time,etc. Now, as per wikipedia, infinitesimal number is a non standard real number whose distance from 0 is less than any positive real number. Therefore it is infinitely close to 0. But, you may notice that this does not make sense until you either divide this infinitesimal with another infinitesimal, which may give you a finite value, or you can do an infinite sum of this very small quantity which may give you a finite value ( an integral ). Commented May 24, 2020 at 3:29
• @Curious This using of infinitesimal is considered a bit sloppy by mathematicians and they instead say that rather than it being a fraction of two non standard small numbers, $\frac{d}{dx}$ ( or $\frac{d}{dt}$ or any other quantity which you are differentiating with respect to) is an operator ( a derivative operator) which will find the derivative of the quantity which is operated upon with respect to the quantity you are differentiating with respect to. Therefore $\frac{dx}{dt}$ will act on position function and give it's derivative with respect to time. Read about derivatives to know more. Commented May 24, 2020 at 3:42

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.