Variables in rectilinear motion

Question

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.

Answer 1

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.

Answer 2

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.