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It states that angle made by acceleration when is between 0 and 89 degrees , acceleration increases whereas in case when it is between 90 - 179 degrees, acceleration decreases.

I didn’t understand why not to include 90 and 180 degree angle ? I am not getting how to draw those components like which reach this solution ?

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  • $\begingroup$ The book leaves the case where $\theta = 90$ and the magnitude of velocity (speed) remains constant but the direction changes for a later discussion, but includes the 180 in the second case. $\endgroup$ Jan 5 at 13:11
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It states that angle made by acceleration when is between 0 and 89 degrees , acceleration increases whereas in case when it is between 90 - 179 degrees, acceleration decreases.

That's not what it says.

The textbook is talking about the instantaneous change in speed (the magnitude of velocity, which is pointing to right in the example), and about how it depends on the angle between the velocity vector and the acceleration vector.

All it's saying is this:

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  • $\begingroup$ Yes.At 90, is it that it turns at variable angle.Like right now it is horizontal.Slowly towards acceleration side , there will be more velocity.It’s like , I am not getting how to make components of angle 90 acc $\endgroup$
    – srijan Sri
    Jan 5 at 13:03
  • $\begingroup$ @Aridhan - I'm not sure exactly what you're confused about (or why you're thinking about components), but at 90 deg, the acceleration vector is vertical - it has no horizontal component (horizontal component is zero). And the vertical component is the same as the whole vector. $a \cos\theta = 0; a \sin\theta = a$ $\endgroup$ Jan 5 at 13:11
  • $\begingroup$ BTW, the fact that the horizontal component (the one parallel with the velocity) is zero is why the speed doesn't change. $\endgroup$ Jan 5 at 13:16
  • $\begingroup$ @Aridhan - is it the phrase "angle made by acceleration" that confuses you? It doesn't mean that an angle was somehow created by acceleration; the phrase "acceleration is making an angle with velocity" is just a way of saying that acceleration and velocity vectors aren't parallel, that there's an angle between them. All possible angles are included (so 90 and 180 as well). The textbook just says that when the angle is 90, the speed is not affected at that instant. $\endgroup$ Jan 5 at 13:24
  • $\begingroup$ Very intuitive answer and good pictures. Can't explain it much simpler than that. @Aridhan, if you still struggle with this, maybe just try to follow the course for a while and get back to this some time later. Your intuition will have increased. If you need more details, there's no way around a standard introductory text on math for physicists. $\endgroup$
    – TBissinger
    Jan 5 at 17:30
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Remember that the direction of the acceleration, when at 90° exactly, is always perpendicular (Normal) to the direction of the velocity. And when the direction of the velocity vector changes, the direction of the acceleration vector changes with it, remaining 90°, Normal, to the velocity vector.

With that, remember that only the tangential component of the acceleration has an impact on the magnitude of the velocity.

As a perpendicular acceleration vector has no impact on the tangential speed, the magnitude of the velocity will stay constant, even though its direction will change.

In your diagram, I think you flipped around the cos and sin. if your acceleration is tangential to your velocity, $a_t = a \cdot 1$ which can only be achieved with $a \cdot \sin(0)$. Meanwhile, the normal acceleration is $a_n = a \cdot \cos(0) = a \cdot 0 = 0$.

In the same way, normal acceleration has no impact on the magnitude of the velocity because $a_t = a \cdot \sin(90) =0$.

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If the force (i.e. acceleration) is perpendicular to the particle's velocity, the absolute value of the velocity will not change but only its direction. Think about a pearl going around on a ring. Its position can be described by a vector pointing from the center of the center of the circle to the position of the pearl on the ring. Its velocity is always perpendicular to that vector, and indeed the distance of the pearl form the center always has the same magnitude.

Therefore, $90^\circ$ is a special case. That works for your text as well: as long as $0^\circ \le \theta < 90^\circ$, there is an increase in velocity, and as soon as $90^\circ < \theta \le 180^\circ$, there is a decrease. All these functions are continuous, so if the change of the magnitude of the velocity changes sign when going from above $90^\circ$ to below $90^\circ$, it must be that this change is exactly $0$ at $90^\circ$.

The $180^\circ$ angle is also included, I think you misread the $\le$ and $<$ signs there.

Edit: Also, because you ask why we don't consider the angle of $90^\circ$: we do. It's just neither an angle at which the absolut velocity dereases nor one at which it increases, so it doesn't appear in the list. This is even said in the note right below the sentence you were asking about.

Edit 2: The relation shown with vectors. We want to know how the absolute value of the velocity changes. Take its square, $|v(t)|^2 = \mathbf{v}(t) \cdot \mathbf{v}(t)$, and calculate the time derivative $$\frac{\textrm{d}}{\textrm{d}t} \mathbf{v}(t) \cdot \mathbf{v}(t) = 2 \mathbf{v}(t) \cdot \left[\frac{\textrm{d}}{\textrm{d}t}{\mathbf{v}}(t)\right] = 2 \mathbf{v}(t) \cdot \mathbf{a}(t).$$ If $\mathbf{v}(t) \cdot \mathbf{a}(t) > 0$, then the absolute value increases, and this is equivalent to $0 \le \theta < 90^\circ$. If $\mathbf{v}(t) \cdot \mathbf{a}(t) < 0$, then the absolute value decreases, and this is equivalent to $90^\circ < \theta \le 180^\circ$. If, finally, $\mathbf{v}(t) \cdot \mathbf{a}(t) = 0$, then the modulus $|v(t)|$ does not change in time, the absolute value of the velocity remains constant. This corresponds to $\theta = 90^\circ$.

Edit 3: I assume from the further questions that you may have problems with a common identity. Take any two vectors $\mathbb{u}$ and $\mathbb{w}$. The angle $\theta_{uw}$ enclosed by these vectors is defined via $$\mathbb{u} \cdot \mathbb{w} = |\mathbb{u}||\mathbb{w}|\cos(\theta_{uw}).$$ This definition can be made for any type of vectors, but especially from three-dimensional real vectors like those describing velocity and acceleration. The angle $\theta$ in your example is the angle formed between $\mathbb{v}(t)$ and $\mathbb{a}(t)$. If you struggle with these concepts, you should pick up any introductory math for physicists book and get familiar with them.

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  • $\begingroup$ Ok.Can you give a more simple example for 90 degree. I am getting confused since I don’t know about centripetal and centrifugal force $\endgroup$
    – srijan Sri
    Jan 5 at 12:03
  • $\begingroup$ That has nothing to do with that, it's just a geometrical picture. It uses an analogy between velocity and acceleration as change of velocity (your question) to position and velocity as change of position (my answer). This specifically avoids talking about centripetal/fugal force. If you wanted to talk about a particle that experiences a constant force at a $90^\circ$ angle, you would need to talk about a particle rotating on a circle with a constant force pulling it toward the center of the circle. $\endgroup$
    – TBissinger
    Jan 5 at 12:07
  • $\begingroup$ Ok.Can you also explain in terms of components actually.I can imagine your answer but in terms of vectors if you .It would be great $\endgroup$
    – srijan Sri
    Jan 5 at 12:13
  • $\begingroup$ I have attached a picture $\endgroup$
    – srijan Sri
    Jan 5 at 12:17
  • $\begingroup$ I put it into an edit. $\endgroup$
    – TBissinger
    Jan 5 at 12:23

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