How force is equal to the product of mass and instantaneous acceleration?

$$\vec{F}=m\vec{a}$$ Suppose that an object of some mass was constantly accelerating ($$\vec{a} \neq 0$$) through sometime interval. With no doubt, the object had an instantaneous acceleration at any instant in that time interval (since the instantaneous acceleration is defind as a limit).

What about the force? How can there occur a force at a specific time (for example, at $$t=2s$$)? Since a specific time is not a time interval the force can spend acting on the object!
What I think of is that forces must spend time intervals acting on objects, so, there can not be a force without a time interval, so how (if there is no force at a specific time) there is an acceleration?

The key is to look at what you mention in your question: it's the idea of the limit. Let's cover what you seem to be familiar with already. When we define "instantaneous velocity as a function of time" to be $$v(t)=\frac{\text d x}{\text d t}$$ what we really mean is that we are looking at the ratio $$\frac{x(t+\tau)-x(t)}{\tau}$$ when $$\tau$$ becomes infinitely small. If we make $$\tau$$ small enough, then it is as if we are looking at the velocity at the instant at time $$t$$, since during the interval $$\tau$$ the velocity is essentially constant. "An instant" is essentially just a really really short time interval.

The above is also true for acceleration ($$a=\frac{\text d v}{\text d t}$$), and thus also true for forces by Newton's second law $$(F=ma)$$. The force "at an instant $$t$$" determines the acceleration "at an instant $$t$$", which tells us how the velocity changes over the interval $$t$$ to $$t+\tau$$ for a sufficiently small $$\tau$$. More specifically for your example, if we have a force applied to an object existing at $$t=2\ \rm s$$, then we can determine what the velocity will be at a time $$(2+\tau)\ \rm s$$.

In other words, just like how when we say $$a(t)$$ we really mean "the acceleration from $$t$$ to $$t+\tau$$", when we say $$F(t)$$ we really mean "the force from $$t$$ to $$t+\tau$$", since force and acceleration are proportional.

• Why the down vote? If I knew what was wrong with my answer I could fix it. – Aaron Stevens Jan 23 at 12:03
• I'm not the downvote but the OP explicitly states they are okay with instantaneous accelerations. They are confused that this implies an instantaneous force, which disagrees with some preconception they have about forces. The kinematics part of this the OP claims to be alright with. – jacob1729 Jan 23 at 14:13
• @jacob1729 Right. I address the force. I start with the known (velocity and acceleration) and then move to the "unknown" talking about forces. It is how you typically teach or explain things to people. I changed some of my wording to help better reflect this. (Also, I like your 1729) – Aaron Stevens Jan 23 at 14:15
• Oh, I missed the last paragraph. I don't know if that fully addresses the OP's concerns or not but I don't think its worth a downvote. I'm not really clear on SE policy on upvoting people you think were unfairly downvoted. – jacob1729 Jan 23 at 14:20
The way I interpret Newton's equation (F = ma) is that, given the details of the forces we know (or believe) the particle is being subjected to, we then try to calculate the acceleration and position. So in fact, we do not know what exactly $$\vec{a}$$ is, as a function of time or whether it is changing, constant, etc... The goal is to determine this.
Consider a simple example of a particle subjected to a constant gravitational force, which is proportional to the particle's mass $$m$$, as per Newton's laws. That is, $$\vec{F} = -mg\vec{e}_z$$, where $$\vec{e}_z$$ is a unit vector directing the force to point in the downwards direction. We then use the equation $$\vec{F} = m \vec{a}$$ to determine that $$\vec{a} = -g \vec{e}_z$$ which tells us that the acceleration is a constant function in time. Note that the force in this case is known to be constant because it doesn't depend on time or where the particle is located. Now consider a case where the particle is subjected to a spring force, with $$\vec{F} = -k \vec{x}$$, so that the force $$\vec{F}$$ is always pointing towards the origin and depends on how far the particle is from the origin. Now, Newton's law tells us that $$m\vec{a}(t) = -k \vec{x}(t).$$ We see that the force is now not constant in time, since it depends on the particle's position, which is changing in response to the force, etc... This constitutes a differential equation, which must be solved given the initial position and velocity of the particle. In this case, it can be done and shows that the particle oscillates back and forth as the force and inertia balance each other, but in general solving these equations is very hard and requires computers or fancy tricks. The main point however, is that in general, neither the force or acceleration are constant in time and understanding this was what lead Newton to (independently) develop the field of calculus.