What I mean is, suppose a ball is fired from a cannon. Suppose the ball is moving at 100 m/s in the first second. Would the ball have started from 1m/s to 2m/s and gradually arrived at 100m/s? And is the change so fast that we are not able to conceive it? Or does the ball actually start its motion at 100m/s as soon as the cannon is fired?

Suppose a 40-wheeler is moving at the speed of 100 kmph. And it collides with a car and does not brake (The car is empty in my hypothesis ;-) ). If a body does not reach a certain speed immediately and only increases speed gradually, does that mean that as soon as the truck collides with the car, the truck momentarily comes to rest? Why I ask this is because the car is not allowed to immediately start at the truck's speed. So when both of them collide, the car must start from 0 to 1kmph to 2kmph and finally reach the truck's speed. Does this not mean that the truck must 'restart' as well? Intuition tells me I'm wrong but I do not know how to explain it physically being an amateur.

When is it possible for a body to accelerate immediately? Don't photons travel at the speed of light from the second they exist?

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    $\begingroup$ You may not be able to concieve the change, but a high-speed camera sure can: youtu.be/_TNSUIsjdpY?t=60 $\endgroup$
    – Rob
    Nov 10, 2016 at 9:09
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    $\begingroup$ If you like this question, you may also enjoy reading this Phys.SE post. $\endgroup$
    – Qmechanic
    Nov 10, 2016 at 11:41
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    $\begingroup$ It probably doesn't fit your definition of "object", but light (photons) starts moving at the speed of light the instant it's created. (i.e. it immediately starts moving at the highest velocity we're aware of.) That's may be a bit of a cheat, as the photon doesn't start "at rest" but starts as not existing. $\endgroup$
    – R.M.
    Nov 10, 2016 at 15:19
  • $\begingroup$ define "object" $\endgroup$ Nov 12, 2016 at 3:54
  • $\begingroup$ @AmbroseSwasey I've used a cannon ball, truck and a car. I've also mentioned photon at the end. If I'm supposed to be careful with the way I use the word object it would be really helpful if you can simply reply to the separate cases where I have removed the ambiguous term 'object'. Cheers :-) $\endgroup$ Nov 12, 2016 at 4:32

6 Answers 6


The answer to your question is "No" as this would require an infinite acceleration and hence an infinite force to be applied.
However to simplify problems it is sometimes convenient to assume an instantaneous jump in velocity as in the left hand graph.

enter image description here

It might be that the actual change in velocity is as per the green line in the right hand graph but because the change in velocity $\Delta t$ takes place over such a short period of time, perhaps $\Delta t \ll 1$ in this example, the approximation to an "instantaneous" change has little bearing on the final outcome.

So every time you see a velocity against tine graph with some sort of "corner", where a gradient (= acceleration) cannot be found, you have to think that the corner is actually rounded but that rounding occurs over a very short period of time compared with the time scale of the whole motion it matters not a lot.

Another example is of problems where a ball rebounds from the ground and you have to find the time it takes to reach a certain height after the rebound.
It is unlikely that you consider the time that the ball is in contact with the ground and slowing down with an acceleration much greater than $g$ (acceleration of free fall) stopping and then initially accelerating upwards at an acceleration greater than $g$.
Usually the sums are done assuming that the acceleration is $g$ all the time because the time that the ball is in contact with the ground is so much smaller than its time of flight through the air.

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    $\begingroup$ Except photons. The get their speed instantly. $\endgroup$
    – talex
    Nov 10, 2016 at 11:37
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    $\begingroup$ @talex I think that is a bit misleading. If photon gets their speed instantly, then it also gets momentun instantly, and then the light source would have to get opposite momentum instantly. I'm not actually sure how the whole thing works, but saying "photons get their speed instantly" doesn't sound the right way to say it... $\endgroup$
    – hyde
    Nov 10, 2016 at 12:34
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    $\begingroup$ momentum is misleading when talking about massless particles :-) $\endgroup$
    – Rory Alsop
    Nov 10, 2016 at 12:43
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    $\begingroup$ @hyde It was kind of bad joke. Photons are born with momentum. Quantum physics is strange :) $\endgroup$
    – talex
    Nov 10, 2016 at 12:51
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    $\begingroup$ @hyde See this question and its answers for the question of "acceleration" in the emission of a photon. $\endgroup$
    – ACuriousMind
    Nov 10, 2016 at 16:52

To answer this, one must understand how motion starts. The motion of the canon-ball starts due to force applied on it. The force directly does not lead to a high velocity. Force leads to acceleration, ie, change in velocity. So, the motion is dictated by

$$F=ma=m\dfrac{\mathrm dv}{\mathrm dt}$$

The change in velocity $\mathrm dv$ during the first moments (which is the question) depends on the infinitesimal time period $\mathrm dt$ you are considering. This time period can be arbitrarily small, and as the force $F$ is fixed, the change in velocity $\mathrm dv$ can be arbitrarily small too. So yes, the motion starts 'gradually', which is to say that the ball doesn't have $0$ velocity and then $100\ \mathrm{m/s}$ without reaching a velocity in between.

Note: We should technically be talking about changes in momentum, instead of velocity. If one adds quantum and relativity to the picture, the analysis becomes more complicated, and the question itself has no meaning in that scenario. I assume you only wanted to know about everyday 'canon-ball' situations.


Nothing propagates instantly

For starters, any real object has a non-zero size. If you apply force to one side of a cannonball, it will start moving before the other side of the cannonball can be affected by that force. The time will be very small ("so fast that we are not able to conceive it") but nonzero. The cannonball will experience some compression, and in general (for reasonable energies) the propagation will be limited by the speed of sound in that material.

So even without even going to gradual acceleration (the main purpose of the cannon barrel - there's a reason why they're so long!), even a simple transfer of momentum by a single impact (e.g one Newtonian ball hitting another) isn't instant.

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    $\begingroup$ Electrons change orbit shells instantly. $\endgroup$ Nov 10, 2016 at 23:09
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    $\begingroup$ "change orbit shells instantly" seems a gross simplification. $\endgroup$ Nov 11, 2016 at 6:53
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    $\begingroup$ Gross simplification: right. The old and new state exist in superposition for a time period, smoothly changing from 100%–0% through to 0%–100%. Any real observation made during that time will find it in one state or the other. $\endgroup$
    – JDługosz
    Nov 11, 2016 at 10:00

In classical mechanics immediately means "immediately", : all the energy transferred at t=0. (I should qualify this as the chosen answer demonstrates, since really t=0 cannot be reached but only approached as a limit, which can be infinitesimally small). A time lag from the firing to the transfer of all the forward momentum of the explosion will only depend on the time distribution of the whole explosive catching fire. This is statistical mechanics and depends on the density, type etc of the material that is being fired.

Explosive materials may be categorized by the speed at which they expand. Materials that detonate (the front of the chemical reaction moves faster through the material than the speed of sound) are said to be "high explosives" and materials that deflagrate are said to be "low explosives". Explosives may also be categorized by their sensitivity. Sensitive materials that can be initiated by a relatively small amount of heat or pressure are primary explosives and materials that are relatively insensitive are secondary or tertiary explosives.



This term is used to describe an explosive phenomenon whereby the decomposition is propagated by an explosive shock wave traversing the explosive material at speeds greater than the speed of sound (340 m/s). The shock front is capable of passing through the high explosive material at supersonic speeds, typically thousands of metres per second.

So there will be a small time lag before the projectile gets the whole front.

If one starts thinking of the quantum mechanical processes that generate the explosion there is an absolute lower limit to the time interval , given by the Heisenberg uncertainty principle in the form of $\Delta(E)\Delta(t)>\hbar/2$. This is in contrast to the classical case. The HUP is fulfilled within the constraints of the classical explosion energy transfers.

  • $\begingroup$ @user157860 the anwer by Peteris applies to your example. Smaller scales maybe, but it is the same. $\endgroup$
    – anna v
    Jul 5, 2017 at 14:17
  • $\begingroup$ @user157860 Only in classical physics. At the impact point quantum mechanics takes over because of the dimensions of "point". The propagation of anything quantum mechanically is limited by the velocity of light c. The dp/dt given to the impact atoms if Feynman diagrams are written for it will be exchanged particles and everything goes with velocity c. $\endgroup$
    – anna v
    Jul 6, 2017 at 4:23
  • $\begingroup$ @user157860 No. only in a universe where there is no quantum mechanics. The underlying level of nature is quantum mechanical, and all classical theories emerge naturally from this layer. It is not the ball velocity, it is the velocity of propagation of signals that is c, that the ball can "understand" it has been hit. $\endgroup$
    – anna v
    Jul 6, 2017 at 4:43

If we want a quick solution to a) how far the ball went, b) what's its exit velocity from the cannon, etc. then we can assume that the ball left the cannon at 100 m/s. Its velocity's x-component can be assumed to be constant, with only the y-component changing due to gravity (e.g., the $v_y = 0$ at the top of the ball's trajectory, but is greater than zero otherwise).

However, if we want to be more picky and realistic, then no the ball does not go from 0 m/s to 100 m/s in 0 s. It may get there quickly, but we must obey ideas such as inertia and impulse. For example, the impulse equation can be written as $F \Delta t = m \Delta v$. If the change in time is zero, then we can't have any change in velocity (meaning going from 0 velocity to some non-zero velocity).


It is certainly true that the acceleration is not instantaneous and occurs over a period of time, as already covered by a number of other answers.

One interesting result of this gradual acceleration that occurs within the barrel that can be observed is that increasing the barrel length results in an increase in the muzzle velocity of the projectile.

For example, this chart of muzzle velocity versus barrel length describes the issue clearly. The bullet is being constantly accelerated by the expanding gases from the deflagrating propellant, and therefore increasing the barrel length (and therefore increasing the time of acceleration) results in a higher muzzle velocity experienced by the bullet.

  • $\begingroup$ I would expect that a really long barrel would actually result in a decreased muzzle velocity when the force supplied by the internal gas pressure is overcome by friction between the projectile and the barrel and the external air pressure. If it were really REALLY long, it might stop entirely. Is that true? $\endgroup$
    – Devsman
    Nov 11, 2016 at 19:22
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    $\begingroup$ @Devsman Probably. There's a balance - a longer barrel within limits means that the gas can apply force to the projectile for longer. Certainly, even in the absence of friction, it would no longer apply any force (or depending on the design of the gun, might apply suction) after a long enough barrel for it to have reached atmospheric pressure. $\endgroup$
    – Random832
    Nov 11, 2016 at 21:24

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