I wrote a post few days earlier on circular motion but it seems i still haven't got the hang of it yet. When is this equation actually true? $s=ut+\frac{1}{2}at^2$

Suppose velocity is given by $v=t-2$ where $t$ is time. If we apply the equation above, it gives $s=0$ meaning no displacement but it surely travelled some distance. Hence showing that this equation is only valid for displacement.

Now in case of circular motion, when using variables along the circle, we take $s$ as the arc length and use the equation above taking $a$ as tangential acceleration. But i just gave a counter-example above that $s$ must be displacement where arc length is distance. How are then we being able to use the equation of motion for curved paths?

Please take it with a grain of salt if you think this is just a roundabout way of phrasing my previous post but this doubt is bugging me out and i think a new thread is needed to solve this issue once and for all.

  • 2
    $\begingroup$ that equation is only valid for constant acceleration, not circular motion $\endgroup$
    – user65081
    Aug 18 at 17:12
  • $\begingroup$ Yes but the function i gave is linear,so acceleration is bound to be constant. $\endgroup$
    – madness
    Aug 18 at 17:27
  • $\begingroup$ I have explained here how to relate that equation with scalers. I suppose you've read that. But I don't know what is your confusion regarding that. $\endgroup$
    – ACB
    Aug 18 at 18:38
  • $\begingroup$ @ACB, i really appreciate the explanation and answer you have provided, but in the above example which I gave, the answer turns out to be $0$ though we used the same definition like acceleration is rate of change of speed,speed is that of distance. I maybe a fool for not understanding that completely. But you deduced that $s=ut+\frac{1}{2}at^2$ works always,here it doesn't. $\endgroup$
    – madness
    Aug 18 at 19:48
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    $\begingroup$ How are you applying $v=t−2$ to the first equation? $\endgroup$ Aug 18 at 20:00

The equation $s=ut+\frac{1}{2}at^2$ applies to motion with constant acceleration in one dimension. We can apply this to curves in space as well, but we need to be careful. To use this equation, $a$ is the acceleration along the path, and $u$ is the initial velocity along the path, but what is $s$? Certainly it can't be displacement, as your circle example shows; the displacement repeats itself around the circle indefinitely, but $s=ut+\frac{1}{2}at^2$ does not repeat itself. It can't also be distance traveled along the path, since this expression can be negative.

So what does this equation tell you? I suppose you could call it a signed distance along the path; I think a better view would be to take the path and imagine "rolling it out" on a straight line. Then $s$ describes the displacement along this line.

In your circle example, you can use this idea to figure out where you are along the circle. Just find $s(t)$ mod $2\pi r$, where $r$ is the radius of the circle; i.e. subtract from $s(t)$ the largest multiple of $2\pi r$ such that the result is still positive, and the result tells you how far along the circle you are from the starting point in the positive direction around the circle.

  • $\begingroup$ Thanks a lot and sorry for the late reply, but as you pointed out that we can roll out the circle and treat it like a line,then shouldn't we be able to make any shape a line?What is the need for displacement then if we can make any shape a line and treat distance like displacement?Also why is it justified to roll out a circle? $\endgroup$
    – madness
    Sep 2 at 10:25
  • $\begingroup$ @madness This would work for any shape. The "rolling out into a line" is just a visual analogy. As long as the acceleration along the path is constant, $s=ut+\frac12at^2$ will tell you the position you are along the path relative to the starting point. $\endgroup$ Sep 2 at 11:57
  • $\begingroup$ Thanks for the kind reply,here is a case.If we throw a ball vertically upward with speed $u$ and $H$ be the maximum height.Now to find the time period,we would use the equation $s=ut-\frac{1}{2}gt^2$,here we use $s=0$ for displacement,so we get $t=2\frac{u}{g}$ as desired,but if we use $s=2H$ as distance,it doesn't give the correct answer. Here the path is a straight line with the fact that we go up once and down once,but according to you we could spread the up and down in both directions and that would give us a straight line of length $2H$ as displacement but it doesn't work. $\endgroup$
    – madness
    Sep 2 at 12:28
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    $\begingroup$ @madness That is why I said "position along the path" rather than "distance traveled". $s=0$ is the correct way to look at that example. And this is why the "stretching the path onto a line" analogy is useful since then the path doesn't double back on itself. Ultimately, the best way to use that equation is only for constant acceleration motion in one direction. What I am described in my answer is a nice mathematical trick to make it work in other scenarios, but in practice I have rarely if ever seen this done. $\endgroup$ Sep 2 at 12:47
  • $\begingroup$ Thanks a lot for the trouble you went through,it cleared most of my doubts,but i still am a dumb kid who is stumbling along the bits and pieces,could you please explain what position along the path is and how $s=0$ is the correct explanation for that case?I would be really indebted to you and it would clear out my doubts about this problem. $\endgroup$
    – madness
    Sep 2 at 13:56

Suppose a particle is being accelerated with constant speed change around a circle. The speed change is $\ddot{s} = c$ for all time, where $c$ is a constant, and $s$ is the distance traveled along the circle. Integrating this twice, we get that the distance traveled at some time $t$ is $s(t) = s_0 + \dot{s}_0 t + \frac{c}{2}t^2$ where the initial time $t_0$ is $0$ and we can say that the particle has traveled no distance yet at time zero, so that $s(t) = \dot{s}_0 t + \frac{c}{2}t^2$.

When the particle has gone full circle, the distance traveled is equal to the circumference: $s = 2 \pi R$, where $R$ is the radius of the circle. You can use this condition to solve for the time it took for the particle to go full circle. Yet the displacement of the particle from where it started is zero.

When the particle is halfway around the circle, its distance traveled is $\pi R$, and its displacement is ${\bf r}(t_{\text{halfway}}) - {\bf r}_0 = 2 R {\bf u}$, where ${\bf u}$ is a unit vector pointing from the starting point towards the antipodal point on the circle.


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