Can the initial time be non-zero?

Question

I am just slightly confused about how initial time is chosen in mechanics. I keep thinking I have understood it but some doubts come up later.

Basically when I derived the constant acceleration equations in a general way I ended up with $v(t)=v(0)+at$ and $x(t)=x(0)+v(0)t+\frac{1}{2} at^2$. However, I didn't set any initial conditions at this point, all I did was derive them using integration then set $t=0$ to get the constants of integration.

Now I understand that we can use a value other than $t=0$ to find the constants, however $t=0$ is convenient.

What I don't understand, is whether by setting $t=0$ to find the constants, I have now somehow implicitly chosen $t=0$ as the initial time? Because if $t=0$ was not the initial time, then it should be impossible to have $t=0$ (am I wrong in thinking this?)

In short, my question is: did I implicitly choose an initial time of $t=0$? If not, then regardless of how ugly the end result becomes, am I free to choose a nonzero initial time so that my first equation becomes $\Delta v = v(0)+at-v(t_0)$ (where I simply subtracted $v(t_0)$ from both sides, where $t_0\ne 0$, as I just said) and the second one becomes $\Delta x = x(0)+v(0)t+\frac{1}{2} at^2-x(t_0)$. I am not concerned with how ugly these are, I am simply wondering if they are correct in principle?

Also, if I wanted to make them not-ugly, I know how to do that: I would choose the $t$ value when finding the constants of integration as $t=t_0$ rather than $t=0$. In other words, my initial time is $t=t_0\ne 0$, hence $v(t)=\int a dt=at+c$ and instead of setting $t=0$ here, I instead let $t=t_0$ to get $v(t_0)=at_0+c$ or $c=v(t_0)-at_0$, hence $v(t)=v(t_0)+a(t-t_0)$. Then I would integrate again to get $x(t)=v(t_0)t+\frac{1}{2} a(t-t_0)^2+c$ and setting $t=t_0$ again gives $c=x(t_0)-v(t_0) t_0$ so that $x(t)-x(t_0)=v(t_0) (t-t_0)+\frac{1}{2} a(t-t_0)^2$.

Answer 1

The formula states that the velocity at Time k1 is equal to the initial velocity at Time k2 + at, where t is k1 - k2. This is because this equation applies only for constant acceleration, so the start time and end time does not matter, but only the duration does.

Answer 2

Your method of derivation is partly correct. There is another method: integrate with lower limit $t=0$ and upper limit $t=t$. Then also you'll arrive at the same equations.

Although you're taking $0$ as initial condition, $t$ in the equation actually means the difference in initial and final time. For example, if you had integrated with lower limit $t_1$ and upper limit $t_2$, in place of $t$, you would have got $t_2 - t_1$. So, $t = t_2 - t_1$.

$t_2$ and $t_1$ may be anything, keeping in mind that $t_2 > t_1$. For example, $t_2$ can be $10s$ and $t_1$ can be $4s$. What you have to put in the equations while solving is $t = t_2 - t_1$. So, any initial time can be taken, not necessarily $0$.

Here, $t_1$ is the initial time and $t_2$ is the final time.

Answer 3

Your graph of velocity (v) against time (T) is a straight line as illustrated below.

The acceleration is the gradient of the graph $a = \dfrac{v_2-v_1}{T_2-T_1}$

But you will see from the graph that the time axis can be relabeled to have the time start anywhere on the time axis.

It is usually the case that time $t$ is the interval $T_2-T_1 = t - 0 = t$ for convenience.

So instead of writing $v_2 = v_1 + a (T_2-T_1)$ one writes $v_2 = v_1 + at$ and the other equations of motion follow on from this eg the area of the trapezium under the graph is the displacement.

Answer 4

What I don't understand, is whether by setting t=0 to find the constants, I have now somehow implicitly chosen t=0 as the initial time? Because if t=0 was not the initial time, then it should be impossible to have t=0 (am I wrong in thinking this?)

It is not impossible to say $t=0$ even if the chosen initial time is larger than zero. Asking for $t=0$ then just corresponds to asking about a value in the past.

You seem to be thinking that there is a physical limitation to an equation like this because we can't measure something before the initial time, and thus using any value of time before the initial time is wrong.

But then remember that your equation is nothing more than a model. Or an approximation. You can use it to predict the future, by plugging in a time way ahead of the current time. You can likewise use it to estimate the past by plugging on a time before the initial time.

Of course, in order to do this, we must first trust the equation to hold true far into the future and far back in the past. If the moving object suddenly changes acceleration in the future or if it did change acceleration in the past, then the equation would not hold. We trust it to hold and know that the results it gives are only true if it does hold.

Mathematically, in order to find the constants of the integration to finish deriving that expression, it doesn't matter at which point in time you plug in some data to calculate the constant. The equation just has to be true in that moment. But the constant is a constant, so it is the same at every point in time - as long as the object does follow this equation.

Answer 5

I have solved the problem. In the process I have found that the book Understanding Physics by Mansfield is much clearer than Tipler and Mosca.

Basically, there seem to be two points of confusion: 1. What does $\int$ really mean? 2. What are the initial conditions?

First of all, what $\int$ really means is $\int_c^x$ where $c$ is an arbitrary constant. This is why we get a constant of integration, because $\int x^2 dx$ is really $\int_c^x x^2 dx = \frac{x^3}{3}-\frac{c^3}{3}$ where $-c^3/3$ is an arbitrary constant, which is why we just let it be $+C$ when we integrate indefinitely. I.e. $\int x^2 dx = \frac{x^3}{3}+C$.

How does this apply to mechanics? In mechanics, the arbitrary constant $c$ cannot be anything. This is because our variable $t$ cannot be anything, since it depends on when we started measuring. For example, I want to find the velocity of a body. If I only started measuring at $t=t_0$, it follows that I have no information about $t<t_0$. This means I can only have a lowest limit of $t_0$. Thus, in physics, it is not appropriate to just say $\int$, we must rather specify the limits: $\int_{t_0}^t$.

So using $a=\frac{dv}{dt}$ we get $\int_{t_0}^t a dt = \int_{t_0}^t \frac{dv}{dt} dt$, ie $v(t)-v(t_0)=a(t-t_0)$ or $v(t)=v(t_0)+a(t-t_0)$, where $t_0$ is not necessarily $0$, as explained above.

At this point, someone clever might say: "You can now let $t=0$ to get $v(0)=v(t_0)+a(-t_0)$, and plugging this back into the equation gives $v(t)=v(0)+at$, which is simpler and neater."

The reason this is wrong, is because he started by saying let $t=0$. As I explained above, since we only started measuring the body (by assumption) from $t=t_0$, it follows that we have no information about $t=0$. Therefore, $v(0)$ is meaningless, so $v(t)=v(0)+at$ is actually un-usable for us, since we don't know $v(0)$: we only started at $t=t_0$! However, if someone were to come and tell us: "I started observing the body before you, and I find that $-t_0$ second before you started measuring, ie at your $t=0$, the body was going at $v(0)=k$", and he tells us an actual numerical value $k$. If this happened, then indeed, $v(t)=v(0)+at=k+at$ is simpler than our equation, and is now usable for us, whereas before it wasn't.

Ok, that clears up the velocity equation. Now we simply integrate it again to get $\int_{t_0}^t v(t)-v(t_0) dt = \int_{t_0}^t a(t-t_0) dt$, ie $x(t)-x(t_0)-v(t_0)(t-t_0)=\frac{1}{2}a(t-t_0)^2$, hence $\Delta x=v(t_0)\Delta t + \frac{1}{2} a(\Delta t)^2$.

This is all general, we still haven't set $x_0$, $v_0$ or $t_0$. So everything is clear: we just go with the convention at this stage and set $t_0=0$, ie we started measuring the body at the same time as we started measuring time. Also, we let $x_0=0$. Therefore, the equations become:

$v(t)=v(0)+at$ and $x(t)=v(0) t+\frac{1}{2} at^2$.