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Recently, I came across a text of a book saying that acceleration acting perpendicular to a velocity only changes the direction of final velocity and not the magnitude. But according to my calculations, I found that the direction changing fact is true but the magnitude also changes simultaneously. This is the way how I did the calculations : According to the calculations I did,I found that Resultant/final velocity has a greater magnitude than the initial one.

Is my calculations correct or is it wrong ? If wrong then please provide me with a specific reason. It would really help me out.

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Well, I am not sure about your physics background so I attempt to explain the issue conceptually without the use of calculus or vectors. For starters, let us clear up what you have done in your "calculations" or rather in your "diagrams." I assume SI units throughout and just motion on a plane (2 dimensions).

  • A particle moves with (say, uniform) initial velocity $u$ along a line,
  • An accleration $a = v$ acts perpendicular to this particle for one second,
  • Thus, the velocity due to acceleration is $v$ along the perpendicular,
  • By the Pythagoras Theorem, the resultant speed is $\sqrt{u^2 + v^2}$.

Now let me highlight some of the conceptual problems with what you've done and clarify them.

  • Acceleration is a continuous process. It does not act only after that one second has passed. If we assume that to be the case then the particle would have already moved by a certain distance and your downward velocity contribution due to the acceleration acts on empty space. This does not make sense.
  • After all, intuitively, if you throw a ball in one direction and hit it with another in the perpendicular, it changes direction. So let us consider the reasonable finite acceleration $a$ first for a very short time $t_0 \ll 1$ (for eg. say $t_0 = 10^{-44}$). In other words, the acceleration acts on the particle at an instant of time.
  • At that exact instance, the velocity $v$ along the perpendicular is $at_0$ but since $t_0 \ll 1$, we have, $v \approx 0 \implies \sqrt{u^2 + v^2} \approx u$. There is a very very small change in direction at this instant.
  • Keep doing this for long enough and your particle would start moving in a circle eventually.

A nice physical example to illustrate this is indeed that of uniform circular motion which can basically be summarised as the converse of your statement. Here, the speed of a particle is constant, but the direction of velocity is changing, which leads to an acceleration always perpendicular to the direction of motion. This also highlights how acceleration and even velocity are continuous in some sense. One can see the cumulative effects of such a perpendicular acceleration in tandem with the continued change in direction here.

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  • $\begingroup$ That is an acceptable answer. Thanks . $\endgroup$ Sep 27, 2022 at 15:24
  • $\begingroup$ @DipanjanDas if you find the answer acceptable, please do be kind and mark it so if possible. Thank you :) $\endgroup$ Sep 28, 2022 at 3:35

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