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According to pythagorean theorem V2 should be greater then V1.But why doesn't it happen?

  • $\begingroup$ What do you get if you calculate $|\vec{v}_2|-|\vec{v}_1|$? $\endgroup$
    – jacob1729
    Oct 23, 2020 at 16:26
  • $\begingroup$ Who do you want to mean? $\endgroup$ Oct 23, 2020 at 16:33

3 Answers 3


It's because the acceleration vector is constantly changing direction so that it is always perpendicular to the velocity vector. Your diagram shows some constant acceleration over some distance/time, but it's key to recognize that that slice of time must be infinitesimally small, since acceleration is not constant. The diagram isn't a great representation of what's happening, since there is no length of time over which you can analyze the instantaneous acceleration as a constant acceleration, since the acceleration vector is changing direction at all times. Averaged over a short period of time, the average acceleration vector will bend slightly left in your diagram, and will not be perfectly perpendicular to the velocity vector. With this proper time-averaged acceleration vector, the velocity vector keeps the same magnitude but changes direction.


You have a side labeled $a.dt\vec j$ with finite length, but $dt$ is infinitesimally small, so the length of that side is also infinitesimally small. This diagram is engaging in a common practice of representing an infinitesimal with a finite quantity. The idea is that this diagram represents the an approximation for a finite value, and we can then use the diagram as aide in evaluating what happens when we take the limit as it goes to zero. So it would be a bit more consistent with standard practice to use $\Delta t$ rather that $dt$.

When engaging in this practice, it's important to keep in mind that we need to do that "take the limit" part, and what survives that depends on the order of each term. $a \cdot \Delta t\vec j$ is first order in terms of $\Delta t$. Remember that the acceleration is the limit as $\Delta t$ goes to zero of $\Delta v/\Delta t$, so when we take a term that is first order in terms of $\Delta t$ (as $\Delta v$ is), and divide it by $\Delta t$, we end up with something zero order in terms of $\Delta t$, i.e., a finite value. That is, we cancel out all the $\Delta t$, so when we take the limit as $\Delta t$ goes to zero, it goes to a finite value.

Now consider the amount by which $v_2$ is greater than $v_1$. I'll set $A = v_2, B = a \cdot \Delta t\vec j, C = v_2$ to make the notation easier. We have $C = \sqrt{A^2+B^2}$, or $C = A\sqrt{1+ \left(\frac B A\right)^2 }$. Using the Taylor series for sqrt, $C = A\left(1+\frac {B^2} {2A^2 }-\frac {B^4} {8A^4 }...\right)$. So the change in the magnitude of the velocity is $C-A = \left(\frac {B^2} {2A^2 }-\frac {B^4} {8A^4 }...\right)$. All of these terms have at least two powers of $\Delta t$, so when we take the limit as $\Delta t$ goes to zero, we get zero (even if we divide by $\Delta t$ to get the rate of this change with respect to t).

Another way of putting it is that the magnitude of the velocity is the square root of the dot product of the velocity with itself: $|v| = \sqrt {v \cdot v}$. If we take the derivative of $v \cdot v$, the product rule gives us $2 v' \cdot v$. The term $v'$ represents the acceleration, and if the acceleration is perpendicular to the velocity, then $v' \cdot v=0$.

Remember that the direction of the acceleration is constantly changing. This diagram presents the acceleration as acting over some time, but given any $\vec a$, the length of time over which the acceleration is exactly equal to $\vec a$ is zero. If you were actually to have the acceleration stay at some fixed value (in terms of both magnitude and direction) over some non-zero time, then the magnitude of the velocity would indeed change.


In the case of uniform circular motion there is no tangential acceleration. In other words, in your diagram $a=0$, and therefore $\Delta v=0$ or $\overrightarrow v_{2}-\overrightarrow v_{1}=0$. The only acceleration is the centripetal acceleration towards the center of the circular motion of radius $r$ and is


Hope this helps.

  • $\begingroup$ The diagram shows acceleration orthogonal to velocity, so it is centripetal, not tangential. $\endgroup$
    – Sandejo
    Oct 23, 2020 at 17:34
  • $\begingroup$ It doesnt help. $\endgroup$ Oct 23, 2020 at 17:34
  • $\begingroup$ I'm not speaking about tangential change.The question was about change. $\endgroup$ Oct 23, 2020 at 17:39
  • $\begingroup$ @AbdullahAlZami OK, then your question is about a change in what? $\endgroup$
    – Bob D
    Oct 23, 2020 at 17:55
  • $\begingroup$ I think you've misread the diagram - the black dot (somewhat confusingly) is not the center of rotation. The acceleration vector is pointing toward the center, so the diagram does not show tangential acceleration. I don't follow your point about the acceleration and change in velocity being zero - the centripetal acceleration is not zero, and the direction of the velocity vector does change. $\endgroup$ Oct 23, 2020 at 18:13

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