# Doubt in Kinematics

I know that this isn't the place for such basic questions, but I didn't find the answer to this anywhere else. It's pretty simple: some particle moves in straight line under constant acceleration from one point $x_0$ to another point $x_1$ during the time interval $\Delta t_1$. When the particle reaches the point $x_1$ it reverses it's movement and goes to another point $x_2$ during another time interval $\Delta t_2$. I want to determine $x_2$, however I don't understand how to do it.

My try was: Let $\Delta x_1 =x_1 -x_0$ be the first displacement and let $\Delta x_2 = x_2 - x_1$ be the second displacement. Then I can calculate two velocities:

$$v_1 = \frac{\Delta x_1}{\Delta t_1}$$

$$v_2 = \frac{\Delta x_2}{\Delta t_2}$$

My thought is then to find the acceleration as:

$$a = \frac{v_2-v_1}{\Delta t_2 + \Delta t_1}$$

But I'm not sure it'll work, since the movement reverses at $x_1$ and since I'm assuming the velocities constant on the intervals.

Can someone help me how to think with this problem, and how to solve it?

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Basic questions are fine here as long as you have tried to find an answer yourself (and it seems you have)! –  Michael Brown Mar 20 '13 at 14:24
What happens to the acceleration at $x_1$. Does it stay the same i.e. the particle is decelerating on it's way back, or does the acceleration switch direction as well or does the particle stop accelerating at $x_1$? –  John Rennie Mar 20 '13 at 14:31
From your text it seems there is a constant acceleration $\ddot{x}_0=:a<0$. At $t=0$ you have a velocity $\dot{x}(t=0)=:v_0$ at the position $x_0$. From time integration it follows that $x(t)=x_0+v_0t+\frac 1 2 a t^2$ and $\dot{x}(t)=v_0+at$. Setting $\dot{x}(t^*)=0$ gives you $x(t^*)=x_1$. From there to $x_2$ should be obvious. –  user9886 Mar 20 '13 at 14:31

From your comment it seems you know $x_0$, $x_1$, $x_2$ (In your question you state you want to know $x_2$), $t_1$ and $t_2$. Now as stated in my comment $$x_1=x_0+v_0t_1+\frac 1 2 a t_1^2$$ and $$x_2=x_1+\frac 1 2 a (t_2-t_1)^2.$$ Solving for $a$ gives $$a=\frac{2 (x_2-x_1)}{(t_2-t_1)^2}.$$ Thus, $$x_1=x_0+v_0t_1+\frac{(x_2-x_1)}{(t_2-t_1)^2} t_1^2.$$ Solving for $v_0$ is all that can be done now. All this can be adapted to whatever exactly is given in your problem.

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Not to nitpick, but I just wanted to point out that your final equation gives x1 whereas the question asked for x2. Someone not reading carefully may not read carefully and become confused. –  David H Mar 20 '13 at 20:47

As you point out you are assuming constant velocities in the intervals, but this does not hold in this case precisely because, as you wrote in the beginning, there is some constant acceleration (rate of change of velocity) in each interval. To do this problem, the key is to note that under constant acceleration $a$ in one dimension, the position $x$ of a particle satisfies $$x(t_2) - x(t_1) = v(t_1)(t_2 - t_1) + \frac{1}{2}a (t_2 - t_1)^2$$ Try applyign this to each interval separately.

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that was my first try. However, I don't have any information about the velocity $v(t_1)$ nor about the acceleration $a$. The only information given are the points $x_0$, $x_1$ and $x_2$ and the time intervals. My try was then to work out the acceleration using this method, but as you pointed out I can't assume the velocities constant on the intervals. What should I do them to work this out ? Thanks very much for your support! –  user1620696 Mar 20 '13 at 14:37
You will probably find user9886's comments useful. –  joshphysics Mar 20 '13 at 15:44

Set up $x_2= x_1+1/2 \times a(t_2-t_1)^2$ as your equation for $x_2$, solve for a, which gives a=2($x_2$−$x_1$)/($t_2$−$t_1$)$^2$, and plug that into the equation $x_2 = x_1 + v_0\times t_1+[(x_2-x_1)/(t_2-t_1)^2]$, and this equation will give you $v_0$ in terms of $x_2$. If these are known coordinates, then you can solve for $v_0$, plug it into the equation for $x_1$, and find v1. then this might help you find $v_2$.

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