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How is acceleration calculated when only the direction changes and the magnitude remains the same?

Guys, I know that my previous examples didn’t make sense and I’m sorry because I’m totally new to this topic and also I don’t think ya’ll actually understand my question.

Acceleration is change in velocity over change in time,right?

So, if the magnitude of velocity changes and the direction is same, acceleration is obviously given a value (for e.g. 2 $m/s ^2$ etc) and the direction of acceleration is the direction of change of magnitude of velocity.

My question is, what if only the direction changes? What if there’s this super advanced object that can change its direction without changing the magnitude of velocity?

How will the acceleration be calculated in that case? How can the magnitude of acceleration have value since there is no change in speed ( magnitude of velocity)?

I’m also aware about Uniform Circular Motion and how acceleration changes constantly even when the magnitude of velocity doesn’t change and how it can be calculated. But uniform circular motion is not an one dimensional direction change.

It will be of huge help if you could explain how acceleration is calculated when magnitude of velocity remains same and only direction of velocity changes in 1D motion (I.e., it goes in the opposite direction and for simplicity let’s assume that this body/object needs very little time to change its velocity from positive to negative) I’m still yet to learn calculus so if these calculations include calculus please do explain it a little in detail. Also, if an object can not change direction without changing speed in 1D motion, then please explain giving an example where the object takes very little time to change direction/ go back. I just want to know how the direction change is shown in acceleration.

Thank you for baring with me and for your time.

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  • $\begingroup$ I dont understand your question. The magnitude of acceleration will just be the force exerted on the ball by the wall divided by the mass of the ball. $a = \frac{F}{m}$ $\endgroup$ – Jdeep Jun 21 '20 at 13:55
  • $\begingroup$ Is that even possible in one dimension, without the acceleration being infinite? When a ball deflects, it slows down and then moves back up with the velocity it had previously, there is always a change in magnitude. The only possible way for a particle to change direction but not change its magnitude is a particle in a circular motion. I've posted a solution for circular motion below. $\endgroup$ – Joshua Pasa Jun 21 '20 at 13:59
  • $\begingroup$ Guys this is not a homework question. In fact it just formulated in my head and my school books don’t even talk explain this stuff so please don’t hesitate to answer. I understand that my previous example had ‘forced acting on it and that confused ppl and made my question senseless but in the new example, instead of a ball, I have used a point object that has potential energy to move. Hope that makes sense $\endgroup$ – Jasmine S Jun 21 '20 at 14:37
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    $\begingroup$ Voting to close as unclear: Your specific question is impossible to answer because 1) An object moving in one dimension cannot change direction without stopping for some instant, 2) an amount of time is required for the velocity to change 3) the velocity function must have continuity in order for the acceleration to be defined. $\endgroup$ – Bill N Jun 21 '20 at 16:00
  • $\begingroup$ BowlOfRed, yes I have viewed that question and the answers there but I think the calculations use calculus so I don’t clearly understand the concept. Maybe if someone would explain the answer in detail, it would be of huge help. Thank you. $\endgroup$ – Jasmine S Jun 22 '20 at 6:31
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Acceleration $a$ is the velocity afterward minus the velocity before divided by the time over which it happened. So $$ a =\frac{ v_{\rm{after}} - v_{\rm{before}}}{t} $$ where $t$ is the time over which the acceleration happened. In this example, $$ a = \frac{ -2 v_\rm{before}}{t} $$.

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Alright, so I think the example you've picked is not the easiest. In simple terms, in 1 dimension you are on a line in which you've chosen a zero position and a positive direction. When an object is moving, if it isn't subject to any force then its speed is always the same. When, however, there is a force acting on the object, this force causes the speed to change. If the force is acting in the opposite direction of the speed's then the speed decreases until it eventually changes its direction and then starts to accelerate because at that point the force would be acting in the same direction of the speed, thus increasing it. This happens over a certain period of time and the average acceleration given to the object is $\Delta v/\Delta t$.

In your example, however, the time over which the force acts on the ball, causing it to decelerate until it changes its direction, is zero. It happens instantly. In this case, talking about forces and acceleration doesn't make too much sense, so we talk about an impulse instead.

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In you example you forgot that velocity has a sign: when the point moves to the right the velocity is positive. When the point is moving to the left the velocity is negative. The acceleration is calculated taking the sign into account.

This causes a serious problem in your example in that you assume an instantaneous change in the direction (sign of the velocity). There you cannot calculate an acceleration: its value would be infinite. Physics does not accept infinities. In reality, all accelerations and velocities have finite values. You first have to slow down your point to reach zero velocity at the position 10, then the sign of velocity changes and the point is accelerated again in the other direction to reach some constant negative velocity. Although you can assume rather high acceleration (e.g. when a ball bounces against a wall) in the real world it is still finite.

There is always an approximation to be taken into account when you calculate physical reality based on mathematical models.

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First, in your example - For an object in $1D$ motion to change direction it will have to come to a halt momentarily.

Editing that.

Once you have a velocity "$v$" in +be direction , only other direction is -ve axis , so final velocity is "$-v$"

Thus the speed of object is constant.

If you know the time interval $t$,

$A = -2v/t$ by first equation of kinematics

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Here is another example. You throw a rock straight up. It is uniformly accelerated downward by gravity. The rock slows to a stop and then moves downward. You want to know how this change of direction works without an infinite acceleration at the moment where it stops.

It might be a little clearer that nothing infinite is needed if you throw the rock in an arc. It follows a smooth parabolic trajectory.

Suppose you threw two rocks side by side just right, one in an arc and the other straight up. They leave at the same height and rise to the same height. In this case, they are at the same height for the entire flight. This should show you that somehow nothing infinite is needed for the straight up and down case.

Acceleration is change in velocity. So let's look at the velocity of the straight up and down case. As the rock starts, it has a large upward velocity. The velocity gets smaller and smaller smoothly, until it reaches $0$ at the top. After the top, velocity starts at $0$ and smoothly gets bigger and bigger downward.

At the peak, velocity is $0$. Just before, it is just slightly upward. Just after, it is just slightly downward. The transition through $0$ is smooth, just like all the other points.

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A better example than the one you chose is the centripetal acceleration.

Velocity is by definition a vector, say $\vec{v}$ and acceleration is a vector too and defined as:

$$\vec{a}=\frac{\text{d}\vec{v}}{\text{d}t}$$

When an object moves with uniform tangential velocity on a circle, the magnitude of the velocity (commonly known as the 'speed') $v=|\vec{v}|$ is constant but the velocity vector changes direction all the time.

In that case the centripetal acceleration $a_c$ is given by:

$$a_c=|\vec{a_c}|=\frac{|\vec{v}|^2}{r}$$

with $r$ the radius of the circle.


Let’s consider a point object along a number line. It starts at 0, moves with a constant speed and reaches point 10 on the number line in 5 seconds. Then without stopping it goes back/reverses to point 6 in 2 seconds.

The venom here is in the 'without stopping'. This means that the change in velocity is instantaneous, i.e. effectuated in $\Delta t=0$ and thus:

$$a=\frac{\Delta v}{0}=+\infty !!$$ But if the change took place in a finite time interval $\Delta t$ then:

the change in velocity equals $\frac{10}{5}+\frac{4}{2}=4$

and the acceleration:

$$a=\frac{4}{\Delta t}$$

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