A point moves from point A to point B. Both points are known, and so is the distance between them.

It starts with a known speed of VA, accelerate (with known constant acceleration a) reaching VX (unknown), then start decelerating (with known constant acceleration -a) until it reach the final point B with a known speed of VB.

So we know : VA, VB, a, and distance between A and B.

How to find VX ?

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We write down some equations, in a semi-mechanical way, and then solve them.
Let $D$ be the (known) total distance travelled.

It is natural to introduce some additional variables. The intuition probably works best if we use time. So let $s$ be the length of time that we accelerated, and $t$ the length of time that we decelerated.

The (average) acceleration is the change in velocity, divided by elapsed time. Thus by looking at the acceleration and deceleration phases separately, we obtain the equations

$$a=\frac{V_X-V_A}{s} \qquad \text{and}\qquad a=\frac{V_X-V_B}{t}\qquad\qquad \text{(Equations 1)}$$

The net displacement while accelerating under constant acceleration is the average velocity times elapsed time. So while accelerating we covered a distance $(1/2)(V_X+V_A)s$. In the same way, we can see that the total distance covered while decelerating is $(1/2)(V_X+V_B)t$. But the total distance covered was $D$. If we multiply by $2$ to clear fractions, we obtain $$(V_X+V_A)s +(V_X+V_B)t=2D \qquad\qquad \text{(Equation 2)}$$

Note that from Equations 1 we have $$s=\frac{V_X-V_A}{a} \qquad \text{and}\qquad t=\frac{V_X-V_A}{a}\qquad\qquad \text{(Equations 3)}$$ Substitute for $s$ and $t$ in Equation 2, and multiply through by $a$ to clear denominators. We obtain $$(V_X+V_A)(V_X-V_A) +(V_X+V_B)(V_X-V_B)=2aD.$$ The left-hand side simplifies to $2V_X^2 -V_A^2-V_B^2$. We conclude that $$2V_X^2=2aD+V_A^2+V_B^2$$ and therefore $$V_X=\sqrt{aD+(V_A^2+V_B^2)/2}.$$

Comment: The algebra turned out to be pretty simple. With other choices of variable, it might have seemed more complicated. An important simplifying device was to choose notation that treats the acceleration phase and the deceleration phase symmetrically. That saved half the work. Moreover, it was the preserved symmetry that made the equations "clean" and easy to work with. Symmetry is your friend.


Hint: You can calculate the time to reach a speed of $V_x$ and a further time to reach $V_B$, so you can find the total distance traveled as a function of $V_x$.

Set this distance equal to $|B-A|$ and solve for $V_x$.

  • $\begingroup$ Tried that, didn't worked, I get an identity. Maybe I made some mistake, my math is not strong enough... $\endgroup$ – St0rM Jul 25 '11 at 17:06
  • $\begingroup$ @StorM: Fine: but why not edit your question to show what you have done so far? $\endgroup$ – Henry Jul 25 '11 at 18:22

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