I was studying the chapter "circular motion" and there I studied the property-"the tangential component of acceleration of a particle undergoing a circular motion just happens to be equal to the instantaneous rate of change of speed of the particle" and it has been proved mathematically in the book. I was thinking about extension of this property to any planar trajectory of a particle.

My Claim-

1)If a particle moves in a plane in any unknown trajectory then at any instant the tangential component of the acceleration will be equal to the instantaneous rate of change of the speed of the particle. I want to know whether my claim is true or not. I have a reasoning in favour of my claim. Though it is not a mathematical one.


2)If we consider random trajectory of the particle,then at any point of the trajectory if we consider the osculating circle(the circle which best approximates the curve around that point) of the trajectory at that point then we can approximate the motion of the particle around that point as the motion of the particle along the osculating circle. So on the basis of this logic I am thinking that my claim is right.

I want to know is my claim right? If yes then is my reasoning a correct explanation for my claim? Can you please provide me a better explanation for my claim? A mathematical explanation is appreciated. Edit:- By tangential component of acceleration I want to mean the component of acceleration along the direction of velocity.

  • 1
    $\begingroup$ Your reasoning is correct but the use of the term 'any unknown' is confusing. 'any given' would be a better expression. $\endgroup$
    – Gert
    Mar 18, 2020 at 16:13

4 Answers 4


If by “tangential acceleration” you mean “the component of the acceleration along the tangent to the trajectory”, your claim is true. Here is the mathematical proof:

$$\frac{dv}{dt}=\frac{d}{dt}(\mathbf{v}\cdot\mathbf{v})^{1/2}= (\mathbf{v}\cdot\mathbf{v})^{-1/2}\,\mathbf{v}\cdot\frac{d\mathbf{v}}{dt}=\mathbf{a}\cdot\frac{\mathbf{v}}{v}=\mathbf{a}\cdot\hat{\mathbf{v}}$$

Since the velocity is directed along the tangent to the trajectory, $\hat{\mathbf{v}}$ is a unit tangent vector and the final expression is thus the tangential component of the acceleration.

  • $\begingroup$ what is meant by a radial trajectory? $\endgroup$ Mar 18, 2020 at 15:56
  • $\begingroup$ One along a straight line through the origin. Just imagine dropping a ball, which accelerates toward the center of the Earth. Its speed is changing but it has no tangential acceleration. $\endgroup$
    – G. Smith
    Mar 18, 2020 at 15:58
  • $\begingroup$ Why is my claim false in the above situation mentioned by you? $\endgroup$ Mar 18, 2020 at 16:01
  • $\begingroup$ I’m going to delete my answer because I think you are using “tangential” to mean “along the tangent” whereas I am using it to mean “along the $\theta$ direction in polar coodinates. $\endgroup$
    – G. Smith
    Mar 18, 2020 at 16:05
  • 1
    $\begingroup$ Actually, I will just edit it to explain why you are correct. $\endgroup$
    – G. Smith
    Mar 18, 2020 at 16:10

Your claim and reason both are correct. Since your claim is a direct implication of the definition of tangential acceleration, looking for a mathematical explanation might not be fruitful.


G. Smith's answer is probably the most concise and precise way to prove that the original post's hypothesis is correct.

But here's another (slow and explicit) way to look at it if it helps.

Start with the original post's image of a "random trajectory" (probably better described as an arbitrary trajectory). At some time $t_0$ the particle is at some location and has some speed $v_0$.

Set up a new Cartesian coordinate system centered on that location whose $x$ axis is aligned tangent to the trajectory at that moment (the $y$ axis will of course be perpendicular to the trajectory). At time $t_0$ the tangential component of the acceleration is $a_x$ and the perpendicular component is $a_y$.

At $t_0$ and in these coordinates the particle has velocity components $v_{x,0}$ and $v_{y,0}$ given by \begin{align} v_{x,0} &= v_0\\ v_{y,0} &= 0. \end{align}

A short time $dt$ later the velocity of the particle (in this same coord system) has components $v_{x,1}$ and $v_{y,1}$ given by \begin{align} v_{x,1} &= v_0 + a_x dt + \mathcal{O}(dt^2)\\ v_{y,1} &= a_y dt + \mathcal{O}(dt^2). \end{align}

What's the speed of the particle at $t_0+dt$?

\begin{align} \text{speed at }t_0+dt &=(v_{x,1}^2 +v_{y,1}^2)^{1/2}\\ &= [(v_0 + a_x dt + \mathcal{O}(dt^2))^2 + (a_y dt + \mathcal{O}(dt^2))^2]^{1/2}\\ &=[v_0^2 + 2 v_0 a_x dt + \mathcal{O}(dt^2)]^{1/2} \\ &=v_0\left[1+ 2 \frac{a_x}{v_0}dt + \mathcal{O}(dt^2)\right]^{1/2}\\ &=v_0 \left(1 + \frac{a_x}{v_0}dt + \mathcal{O}(dt^2)\right)\\ &=v_0 + a_x dt + \mathcal{O}(dt^2). \end{align} To first order in time only the tangential acceleration changes the speed.

To be explicit, the time derivative in the speed is the limit, as $dt$ goes to zero, of the difference between the speed at $t_0+dt$ and $t_0$, divided by $dt$:

\begin{align} \frac{d\text{(speed)}}{dt} &= \lim_{dt\rightarrow 0} \frac{[v_0 + a_x dt + \mathcal{O}(dt^2)] - v_0}{dt} \\ &= a_x. \end{align}

i.e. the time derivative of the speed is the tangential acceleration.

  • $\begingroup$ Can you please tell what is that O(dt)^2 $\endgroup$ Mar 18, 2020 at 19:27
  • $\begingroup$ It's a shorthand for terms that include $dt^2$ (and higher powers). Like for a function $f(x)=A+Bx+Cx^2+Dx^3$ you could write $f(x)=A+Bx+\mathcal{O}(x^2)$. It's a very convenient notation if you don't care about the exact form of these "higher-order terms". $\endgroup$
    – Alex
    Mar 19, 2020 at 19:24

Acceleration along (parallel to) the direction of motion changes the speed, but not the direction.

Acceleration perpendicular to the direction of motion changes the direction of motion, but not the speed.

Starting with the velocity equation, $$\vec v=\vec v_0+\vec a \Delta t$$ If the velocity and acceleration are parallel, then this equation can be reduced to one dimension. $$v=v_0+a\Delta t$$ The one-dimensional component of the velocity is changing in magnitude, but not direction.

This is true if the time interval $\Delta t$ is finite or infinitesimal $dt$.

For the perpendicular case, the direction is changing all the time, so we'll use a very small (infinitesimal) time interval. $$\Delta t \rightarrow dt$$ We can square both side of our equation $$v^2=(\vec v_0+\vec a t)^2$$ When we expand out the right side of this equation, we get $$(\vec v_0+\vec a dt)^2=v_0^2+(\vec v_0 \cdot \vec a )dt+ \vec a \cdot \vec a (dt)^2$$ As typical in calculus, we drop the $(dt)^2$ term, (because it is "extra small"). $$v^2=v_0^2+(\vec v_0 \cdot \vec a )dt$$ If $\vec v \perp \vec a$, then $\vec v \cdot \vec a=0$.

Which gives us $$v^2=v_0^2$$ and so the magnitudes are the same. $$v=v_0$$ And since the speed is the magnitude of the velocity, the speed stays the same when the acceleration is perpendicular.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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