There is this square of side length $a$. It's opposite vertices are being pulled in opposite directions with a constant velocity $u$. The question here is what is the velocity of the remaining pair of opposite vertices?

I tried to use the rules of constrained motion and made the velocity of each point along one of the edges equal as there is no relative velocity along that line always. I ended up getting that the velocity of the remaining two vertices would also be $u$. But I then reminded myself that I completely neglected the role of the velocity of the opposite vertex (whose velocity is initially given) in calculating the velocity of the other pair of vertices. Then I tried to use center of mass concept as the center of mass is stationary but was confused how to plug in the various values.

I'm really confused and want somebody to help me out of this question. Thanks.

  • $\begingroup$ This might be better suited to Mathematics, since homework-like problems are not well-regarded on this board and this problem is really more about geometry and implicit differentiation rather than about physics. $\endgroup$ – Michael Seifert Aug 12 at 13:46
  • $\begingroup$ I agree this is mostly a geometry problem (like all kinematics problems are). $\endgroup$ – ja72 Aug 12 at 17:24

Here is how to solve problems:

  1. Always draw a sketch first that includes all relevant information.


  2. Find what is invariant (remains constant) and what may be symmetric. Here the vertices of the square move along the diagonal lines.
  3. Rephrase the problem in terms of the stated symmetries. Here pretend the diagonals are immovable walls that are 90° apart and draw only one side as it slides against the walls. One end with $u$ and the other with $v$.
  4. Use geometry (hint similar triangles) to find $v$ as a function of $u$.


  • $\begingroup$ Thanks. Bit 3 was really beautiful. I figured it out! $\endgroup$ – Gourav Mahunta Aug 12 at 18:36
  • $\begingroup$ Step 4 should be v as a function of u ! $\endgroup$ – Eli Aug 12 at 21:30
  • $\begingroup$ @Eli - yep. fixed. $\endgroup$ – ja72 Aug 12 at 21:37

Perhaps this help you to solve the problem

enter image description here

thus $\begin{aligned}2u=\dfrac {\Delta S_{1}}{\Delta t}\\ 2v=\dfrac {\Delta S_{2}}{\Delta t}\end{aligned}$

$\begin{aligned}\Rightarrow \\ v=\dfrac {\Delta S_{2}}{\Delta S_{1}}u\end{aligned}$


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