# Is $dv/dt$ always zero when dealing with 'limiting velocity'?

I am confused on one of my practice problems. When dealing with 'limiting velocity', is $dv/dt$ always zero?

• Please link to use of 'limiting velocity'. – Qmechanic Jul 3 '17 at 6:51

I'm assuming limiting velocity means the same as terminal velocity, in which case the answer to your question is yes. The terminal velocity is defined as the point during free fall with drag (or whatever other process where there is a similar drag force) at which the force due to gravity is the same as the force from the drag.

In the language of Newton's laws, this implies that the total force acting on the body is balanced, $\textbf{F}=0$, and so $\textbf{a}=\mathrm{d}\textbf{v}/\mathrm{d}t=0$.

Assuming limiting velocity is terminal velocity: no, $dv/dt$ (or acceleration) is not necessarily always zero.

However, $dv/dt$ becomes (or tends to) zero after a certain amount of time $t$, and stays zero for any time greater than $t$ (assuming there are no changes in the forces being imparted on the object in question after terminal velocity is reached).

### Examples

1. Accelerating a car from rest all the way to its top speed (its top speed is its terminal velocity). That you could get the car up to top speed from rest implies that acceleration was non-zero once. Otherwise would mean you would be sitting in a parked car.

2. Skydiving from a plane cruising at a constant altitude. Depending on how one orients their body during free fall, terminal velocity can range between 200km/h - 300km/h. That you could get up to that speed implies that you should experience some acceleration during descent. Otherwise would mean you could simply jump out and "fly" along with the plane on its flight path.

The important bit is that you experience constant velocity (no acceleration) after some time.

Yes, dv/dt = 0 at limiting/terminal velocity.

If your initial and final velocities differ though, it's a bit less simple than that, since net acceleration is non-zero when there is a change in velocity.

To demonstrate this point

Let's say your limiting velocity is 70m/s, and your initial velocity is also 70m/s (in the same direction).

No acceleration is required to reach the limiting velocity since you are not changing velocity, therefore your initial, net, and final acceleration is 0m/s^2.

Now let's say that your limiting velocity is 70m/s, but your initial velocity is 30m/s (in the same direction).

You accelerate, for argument's sake, to limiting velocity in 8 seconds. If you started accelerating at a constant rate from the beginning instant of the scenario, your initial acceleration would be 5m/s^2. Your final acceleration would be 0m/s^2.

If you end the scenario the instant you reach limiting velocity, then with a change in velocity (or dv) of (70 - 30), your net acceleration for the entire scenario would be:

a(net) = dv/dt

a(net) = 40/8

a(net) = 5m/s^2 (the same as your initial acceleration)

In the above scenario, however, if you were to remain at terminal velocity for, say, an additional 2 seconds after, then your net acceleration would not be the same as your initial acceleration, assuming you include time of non-acceleration (you wouldn't if you're calculating only for the time spent accelerating).

You would have a dv of (70 - 30 = 40) as before, but the dt of the entire scenario would change from 8s to 10s, resulting in:

a(net) = dv/dt

a(net) = 40/10

a(net) = 4m/s^2.

Even though your initial acceleration remains as 5m/s^2, and your final acceleration remains as 0m/s^2, your net acceleration is now 4m/s^2.

To Conclude

At limiting velocity, dv/dt = 0, since a = 0 and a = dv/dt. This does not necessarily mean that a(net) nor a(initial) nor any other instance of a is also equal to 0.

• The average acceleration (what you call "net") is not given by a derivative. Only the instantaneous quantities can be derivatives. So the discussion about the average values is not really relevant to the question about $\frac{dv}{dt}$. – nasu Jul 3 '17 at 13:43