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I was playing around with the equation in the title when I realized it isn't as simple as I thought. I basically changed the parameters in the derivation given in the book because I was wondering why $t=0$ in particular had to be the starting point. The result was the following, which I am unable to explain.

If we let the times be $t_1$ and $t_2$ instead of $0$ and $t$, then we get $\Delta x = v_{av} \Delta t=(\frac{v_a+v_b}{2})(t_2-t_1)$ where $a,b$ are arbitrary. This leads to the weird result $\Delta x = (\frac{v_0+at_a+v_0+at_b}{2})(t_2-t_1)=(v_0+\frac{1}{2}a(t_a+t_b))(t_2-t_1)=v_0\Delta t+\frac{1}{2} a(t_a+t_b)\Delta t$. *

My question is, is this a correct result? If so, how useful is it? If you replace $a,b$ by $1,2$ it becomes the general case of $\Delta x = v_0 t+\frac{1}{2} at^2$.

* $v_{av}$ means average velocity

Raghib
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  • If we're talking about an object with constant acceleration, then just saying $v_{av}$ is the average velocity isn't enough; the average velocity depends on the interval of time. You need to specify over which interval of time $v_{av}$ is an average. – Kajelad Dec 31 '17 at 23:31

1 Answers1

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$a$ and $b$ can't be arbitrary. The equation $$v_{av} = \frac{\Delta x}{\Delta t} = \frac{x_2-x_1}{t_2-t_1}$$ implies (under the assumption that $a=\frac{dv}{dt}$ is constant) that $v_{av} = \frac{v_1 + v_2}{2}$.

If $v_a = v(t_a)$ and $v_b=v(t_b)$ were allowed to be arbitrary, then $v_{av}$ would not be well-defined unless $v$ happened to be constant. \

So the way to derive the correct equation starting from $\Delta x = v_{av}\Delta t$ is $$\Delta x = v_{av}\Delta t = \frac{v_1 + v_2}{2}\Delta t = \frac{2v_1 + a\Delta t}{2}\Delta t = v_1\Delta t + \frac 12a(\Delta t)^2$$

where $\Delta t = t_2 - t_1$.

To confirm our result, we can simply derive the correct equation from first principles. Consider a particle undergoing constant acceleration over the time interval $[t_1, t_2]$. Then, from the definition of acceleration, we have $$a = \frac{dv}{dt}$$ and thus $$\int_{t_1}^{t}a dt' = \int_{t_1}^{t} \frac{dv}{dt'}dt' \\ a\int_{t_1}^{t}dt' = \int_{v(t_1)}^{v(t)}dv \\ a(t-t_1) = v(t) - v(t_1)$$ where $t\in [t_1, t_2]$. Then, integrating again, we get $$\int_{t_1}^{t} a(t' - t_1)dt' = \int_{t_1}^t [v(t')-v(t_1)]dt' \\ a\left[\frac{t^2-t_1^2}{2}-t_1(t-t_1)\right] = \int_{t_1}^{t} \frac{dx}{dt'}dt' - v(t_1)(t-t_1) \\ a\left[\frac 12(t-t_1)^2\right] = x(t)-x(t_1) - v(t_1)(t-t_1)$$ Then setting $t=t_2$ and solving for $\Delta x = x(t_2)-x(t_1)$, we get $$\Delta x = v(t_1)\Delta t+\frac 12a(\Delta t)^2$$ where $\Delta t = t_2-t_1$.

user517641
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  • Thanks for your answer. While I understand your method, I tried getting it from $\Delta x = \int_{t_1}^{t_2} v dt$ and got a different answer, could you please point out where I went wrong? : $\int_{t_1}^{t_2} v dt = \int_{t_1}^{t_2} v_0+at dt = v_0t_2+\frac{1}{2}a t_2^2-v_0t_1-\frac{1}{2}a t_1^2=v_0 \Delta t + \frac{1}{2} a \Delta t(t_1+t_2)$. – Raghib Jan 01 '18 at 09:51
  • $v\ne v_0+at$. Remember that $t_1$ is your starting time so you have to express any duration with respect to it. I.e. it should be $$v=v_0 +a (t-t_1)$$ – user517641 Jan 01 '18 at 13:05
  • But if we let $t=0$ in that equation we get $v(0)=v_0-at_1$ so it becomes $v=v(0)+at$. – Raghib Jan 01 '18 at 15:57
  • If we're considering the interval $[t_1, t_2]$, then $t=0$ doesn't have any special importance. If you want to set $t$ to the beginning time, it should be $t=t_1$. Then you get $v(t_1) = v_0$, where $v_0$ just means the initial speed not the speed at $t=0$ (but because of the possible confusion I'd prefer to use $v_1$ in that equation). – user517641 Jan 01 '18 at 17:29
  • So the reason my equation is wrong is because I used constants as the limits? So if I let the limits be $c,t$ where $c$ is constant and $t$ is variable it becomes $\Delta x = \int_c^t v(0)+at dt = v(0)t+\frac{1}{2}a t^2+C$ for some constant $C$. Thus letting $t=0$ we get $x(0)-x_0=C$ and substituting this in gives $x=x(0)+v(0)t+\frac{1}{2}a t^2$. Is this correct now? Then if we want to let $x(0)=0$ we can simply say $x_0=0$ when $t_0=0$ (this is the same as saying $x(0)=0$), and this gives the standard equation of $x=v(0)t + \frac{1}{2} a t^2$. – Raghib Jan 01 '18 at 17:34
  • No. You can absolutely use constants as your limits of integration, but you should use $v=v_1 + a(t-t_1)$ and NOT $v=v_0+at$ which is wrong if our time interval starts at $t=t_1$ instead of $t=0$. Setting up the integral as you did, you should get $$\int_{t_1}^{t_2}vdt = \int_{t_1}^{t_2}[v_1+a(t-t_1)]dt = v_1(t_2-t_1)+a\left[\frac{t_2^2-t_1^2}{2}-t_1(t_2-t_1)\right]$$ Simplify that to get the result. – user517641 Jan 01 '18 at 17:36
  • If you're still not understanding why $v=v_1 + a(t-t_1)$ in general, can you tell me where the usual equation $v=v_0+at$ comes from? How is it derived and what assumptions are made in its derivation? Perhaps we can nail down why you're trying to use that equation in a situation where it doesn't apply. – user517641 Jan 01 '18 at 17:44
  • Here is what I did. I used the equation you gave $v=v_1+a(t-t_1)$. I then set $t=0$ to get $v(0)=v_1+a(-t_1)$. Then I plugged this back into $v=v_1+a(t-t_1)$ to get $v=v_1+at+a(-t_1)=at+(v_1+a(-t_1))=v(0)+at$. Therefore, your equation is equivalent to $v=v(0)+at$, so I am not seeing why I can't let $v=v(0)+at$. Sorry if I am still misunderstanding. – Raghib Jan 01 '18 at 17:56
  • Oh, I see. You're trying to derive the equation $x=v(0)t+\frac 12at^2$ from the more general result. I thought you were getting confused by the notation. Then your method does definitely work, but a simpler method -- knowing $\Delta x = v(t_1)\Delta t + \frac 12a(\Delta t)^2$ is valid over any time interval $[t_1, t_2]$ -- is just to take $t_1=0$ and $x(t_1)=0$ to be your starting position, plug those values in to the more general equation, and immediately derive the first one. – user517641 Jan 01 '18 at 18:05
  • Thank you for your help. Could you please look at my new question? I thihk I have expressed my difficulties better there: https://math.stackexchange.com/questions/2587860/is-my-derivation-of-constant-acceleration-equations-correct I will close both questions by accepting your answer here and the answer I get on the other question as soon as someone answers it, since they are linked together. – Raghib Jan 01 '18 at 18:08