About Second equation of motion

[rudransh verma]

I was wondering how the second equation of motion produces negative displacements $s= ut+\frac 12 at^2$ . Is $\frac12 at^2$ kind of distance operator?

 

[DrClaude]

Not sure what you mean by "distance operator", but I think that it is easiest to think of the equation as
$$
s = (u + \frac{1}{2} a t) t
$$
where the parenthesis represents the velocity at time $t$. You will get negative displacements whenever that velocity is negative.

 

[jbriggs444]

I was wondering how the second equation of motion produces negative displacements $s= ut+\frac 12 at^2$ . Is $\frac12 at^2$ kind of distance operator?

You can get a negative displacement if $u$ (the initial velocity) is negative.

You can get a negative displacement if $a$ (the ongoing acceleration) is negative and you wait long enough (or look far enough into the past).

You can get a negative displacement if $t$ is negative and $u$ is positive. So you are looking at displacement at some time in the past.

Graphically, you are looking at a parabola for which some portion extends below the t axis.

The way I read the second equation of motion is that $s = ut$ (Current displacement is equal to initial velocity times elapsed time). But if there is a constant acceleration, $a$, then that results in an additional displacement of $\frac{1}{2}at^2$ that simply adds on.

As I understand @DrClaude, he takes the viewpoint that for constant acceleration, the average velocity is attained halfway through the time interval, so it is given by $u + \frac{1}{2}at$. Multiply average velocity by the total time interval $t$ to get $s=(u + \frac{1}{2}at)t$.

Six of one, half dozen of the other.

 

[rudransh verma]

As I understand @DrClaude, he takes the viewpoint that for constant acceleration, the average velocity is attained halfway through the time interval, so it is given by u+12at. Multiply average velocity by the total time interval t to get s=(u+12at)t.

So $u+\frac12at$ is average velocity and when this is -ve then s is also -ve. The eqn of $s=ut+\frac12at^2$ is same as ${x_2}-{x_1}= {v_{avg}}t$.
u is the initial velocity and due to a acceleration somewhere in time interval t ${v_{avg}}$ is achieved. v is the final velocity not included in the eqn $u+\frac12at$

 

[jbriggs444]

So $u+\frac12at$ is average velocity and when this is -ve then s is also -ve.

Yes -- if we take $t$ as positive. But the equation still holds for $t$ negative if the constant acceleration was constant in the past.

u is the initial velocity and due to a acceleration somewhere in time interval t ${v_{avg}}$ is achieved.

Here I am not sure whether there is a language difficulty or whether you are contemplating a non-constant [1 dimensional] velocity and invoking the mean value theorem. Yes, under these conditions, I believe that the mean value theorem guarantees that the average velocity be attained as an instantaneous velocity somewhere within the time interval.

We normally restrict ourselves to constant accelerations when using the SUVAT equations so that instantaneous velocity is equal to average velocity exactly at the midpoint of the interval.

 

[rudransh verma]

Here I am not sure whether there is a language difficulty or whether you are contemplating a non-constant but continuous acceleration and invoking the mean value theorem.

Constant acceleration. Yes!