Suppose an object is moving with varying acceleration in time. What does it mean when it hits a point where $\frac{dv}{dx}=0$?

Does it mean the object has hit maximum velocity?
Assume the object starts from rest and then ends its journey at a rest position.

  • 1
    $\begingroup$ This is an elementary calculus question. No need to involve any physics at all! $\endgroup$
    – D. Halsey
    Commented Sep 6, 2020 at 18:17

5 Answers 5


At any instant $\frac{d v(t)}{d x(t)}$ is the same as $\frac{d (v(t)}{dt}\cdot \frac{dt}{d x(t)}=\frac{a(t)}{v(t)}$. So it's the ratio of acceleration and velocity at the instant.

It's zero when the acceleration is 0. An acceleration of zero at an instant either implies a local minimum or maximum for the velocity, or a point of inflexion, or a uniform velocity motion (no force)


I am taking v to be a continuos and differentiable function of x$$v=f(x)$$


$$\frac {dv}{dx} = 0 $$ means either v is maximum or minimum at this value of x. This is following simple Calculus.

Without any additional information about its function, it is difficult to say anything else.

If the motion is in one direction only (moving forward only), then it means maximum.

But if it follows a complex trajectory, then it should perhaps represent both maximum and minimum.

You can further check whether it is maximum or minimum with simple Calculus as follows :

For maximum,$$\frac {d^2v}{dx^2} < 0$$

For minimum,$$\frac {d^2v}{dx^2} > 0$$

In case $$\frac {d^2v}{dx^2} = 0,$$ then it is neither minimum nor maximum.


When you use the expression $\frac{dx}{dy}$ you mainly investigate the change of the variable $x$ with respect to variable $y$.

The expression you mentioned:

$$\frac{dv}{dx} = 0$$

indicates that the velocity is stable (could be equal to zero or any other value) since there is no changing of the velocity in that position. Same could be applied to the acceleration expression:

$$v\frac{dv}{dx} = 0$$

In this case the acceleration equals to zero and hence no change in the velocity at that certain position.


The fact that $\frac{dv}{dx}=0$ means the difference in velocity between two infinitesimal nearby points in the $x$-direction is zero. Obviously.
What does this mean physically? Well, if the velocity doesn't change when moving on the trajectory of a particle (or any other object) to nearby points, the velocity is constant in the x-direction. For a given time. So at that given time, the velocity is constant.


The $ \frac{dv}{dx}=0$ suggests that you are either moving with constant velocity, the object has become stationary or you're going to turn.


1.You are moving with constant velocity along a line. Since the velocity is constant with respect to distance, the derivative is zero.

2.Becoming stationery: A body starting with velocity $v$ on a surface being retarded by friction, so at the final point of motion, the $\frac{dv}{dx}=0$

3.Changing direction: After the projectile is at top of the parabola ( vertical component of velocity), a spring-block oscillator with a block at maximum extension.

  • $\begingroup$ Delta v is zero, not v. $\endgroup$
    – rghome
    Commented Oct 8, 2020 at 9:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.