If I apply a force on an object at rest and it’s velocity is constant (after force is applied) wouldn’t that mean that the acceleration is 0? Then how would $F=Ma$ make sense? My thinking here is that change in position means change in velocity but that doesn’t necessarily mean that there is a change in acceleration.

  • $\begingroup$ Welcome to SE. Please edit your title. "FMA" isn't a known acronym, so this title won't help the search tool on the site. $\endgroup$
    – Miyase
    May 23 at 16:24
  • $\begingroup$ If it is at rest and then starts moving, the average acceleration is clearly not zero. Although, at any point in time, the instantaneous acceleration might be zero. $\endgroup$
    – hft
    May 23 at 19:06
  • $\begingroup$ Your own words: "after force is applied" explain it. The acceleration is only zero after force is being applied. Not while it is being applied. $\endgroup$
    – Steeven
    May 24 at 12:22

7 Answers 7


It is normal that an object has a constant velocity while we apply a constant force. But it is an indication that there is another force acting in the opposite direction, usually some friction force.

In the expression $F = ma$, the $F$ is not an applied force, but the net force on the object. If the net force is zero, the acceleration is zero and the velocity is constant.


$a$ is the rate of change of velocity. So if $a$ is zero, $v$ is constant and doesn't change over time.

So while you're applying a force, the velocity is changing and $F=ma$. After the force stops, the acceleration is 0, so the velocity stays the same


My thinking here is that change in position means change in velocity

This is wrong. A change in position does not necessarily mean a change in velocity. If an object is moving at a constant velocity its position is changing. A change in position only means a non-zero velocity, not necessarily a changing velocity.

but that doesn’t necessarily mean that there is a change in acceleration.

Similarly, a change in velocity implies a non-zero acceleration, not necessarily a change in acceleration.

You might also be getting confused about average acceleration vs instantaneous acceleration.

Clearly, in your example, there is an non-zero average acceleration since the velocity changed. This implies that at some point in time there was a non-zero instantaneous acceleration, but not necessarily at all times. In fact, the instantaneous acceleration is only non-zero during the time that the net force is being applied. Before the net force is applied (when the mass is at rest) or after the net force is applied (when the mass is moving at a constant velocity) the instantaneous acceleration is zero.

It may help for you to plot the position as a function of time. Then plot the slope of the position as a function of time (this is the velocity). Then plot the slope of the velocity as a function of time (this is the acceleration).


If I apply a force on an object at rest and it’s velocity is constant

This is not possible. Newton's first law already says that if you apply force, velocity will change. So the situation you describe cannot occur.

If you did apply force and velocity did not change, then the net force must also somehow be zero. Perhaps for example you are pushing the object into a wall, and we can say the wall must be exerting an equal and opposite force which cancels you out. Note that what matters is net force.

  • $\begingroup$ @GiorgioP No reference to action–reaction pairs is made in the answer, only the fact that the person→block and wall→block forces happen to be equal and opposite (not necessarily as an immediate application of Newton's third law). $\endgroup$
    – Jivan Pal
    May 24 at 10:35
  • $\begingroup$ You're right. Sorry, I misunderstood what was written. I am going to cancel my comment. $\endgroup$
    – GiorgioP
    May 24 at 10:53

There's two things to note here. The first one is that both quantities $\mathbf{F}$ and $\mathbf{a}$ are actually functions of time, not just numbers. If one wants to be explicit, one can write this on a value-by-value basis as

$$\mathbf{F}(t) = m \cdot \mathbf{a}(t)$$

but the notation

$$\mathbf{F} = m\mathbf{a}$$

also makes sense provided you suitably "overload" the multiplication operator to take a function as one input, which is then defined pointwise. As a result of this, though, one way your situation can occur is if the force was only nonzero at a certain point in time. If it is later zero, the object need not keep accelerating. There is only acceleration so long as force is present at the same instant of time.

The second possibility concerns the more precise meaning of $\mathbf{F}$. It is not just "any" force, but rather it is the total of all forces acting on the object:

$$\mathbf{F} = \sum_{i \in I} \mathbf{F}_i$$

and thus another way your situation can occur (though not at all mutually exclusive with the first) is for one force to cancel another out. This is what happens when pushing a wheelbarrow at constant speed. The rolling friction is exactly counteracting the force from your hands.


Newton's laws of motion are the foundation of classical mechanics. Newton's second law states that, ''The rate of change in momentum of a body is directly proportional to the force imposed on it. The direction of change in momentum and the force imposed are indifferent'' and its mathematical expression is $F = ma$ where $F$ is the applied force, $m$ is the mass of a body and $a$ is acceleration or rate of change in velocity. If $a = 0$, there is no change in velocity over time which means no force is applied.

My thinking here is that change in position means change in velocity but that doesn’t necessarily mean that there is a change in acceleration.

It is wrong. Acceleration is defined as change in velocity over time. And also there is no change in acceleration i.e acceleration is constant while force is applied but there is change in momentum. See the derivation of $F = ma$ here for better understanding.

So while a force is applied on a body there must be a change in velocity as time changes or acceleration. Note that the net force applied on a body shouldn't be zero or there will be no change in velocity. After the force stops working, the force on a body becomes zero and so velocity remains constant [According to the first law , there is no change in inertia of a body if there is no force working on it].


If I apply a force on an object at rest and it’s velocity is constant (after force is applied) wouldn’t that mean that the acceleration is 0?

After the force is applied, i.e. when the force is no longer being applied, sure, that's right: No force means no acceleration, which in turn means constant speed/velocity, whether that is zero speed (i.e. stationary) or non-zero (i.e. moving).

Then how would $F=ma$ make sense?

This law tells you that the acceleration $a$ of a body at any particular point in time is proportional to the net force $F$ being exerted on it at that same point in time. Whilst you are pushing, you are exerting a force, and thus the object is accelerating. When you stop pushing, you are no longer exerting a force, and thus the object is no longer accelerating, though it very well may be moving (i.e. have non-zero speed/velocity).

An example involving a car:

When the car is stationary, its velocity is zero. When your foot is not on the gas/accelerator pedal, the force that the engine exerts on the car is zero, and thus the car's acceleration is also zero, meaning its velocity is not changing; the velocity stays at zero, i.e. the car remains stationary.

As soon as you press down on the pedal, the engine starts exerting a force on the car, meaning the car starts to speed up. If you hold your foot down on the pedal a certain amount, the force will be constant, meaning the acceleration will be constant, meaning that the rate at which the car's speed increases is also constant (assuming there is no friction between the car's tyres and the road). That is, suppose the car's mass is 1000kg, and the force exerted by the engine is 3000N. This means that the car's acceleration is 3000N/1000kg = 3 meters per second, per second. What this means is that after 1 second, the car's speed is 3m/s; and after another second has passed, it is 6/ms; and after another second, it is 9m/s; and so on — as long as your foot remains on the pedal that same amount, the car will continue to speed up at that same constant rate. That is what is meant by constant acceleration.

Release your foot from the pedal and the car will stop accelerating, so e.g. if you released the pedal after 3 seconds, the car's speed would be 9m/s, and it would remain at 9m/s until some other force acted on the car. In other words, the car would coast at 9m/s, despite no force acting on it. An example of such a subsequent force might be when you press the brake pedal, which would apply a force opposing the direction of motion, resulting in an acceleration opposing the direction of motion (a.k.a. a deceleration), resulting in the car slowing down to an eventual stop.

The only reason the car doesn't continue to coast indefinitely without application of the brakes is that there is indeed friction between the car and the road. Such friction acts in the same way as a braking force, but it is a much smaller force, so it takes much longer for the car to come to a complete stop.


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