Ants on a tetrahedron

Okay, I have a doubt. Please follow my approach and tell me where I am wrong. First, find the velocity of approach of A and B that is velocity along the line joining the two. That is 3/2, since these are equilateral triangle. Since at every instant the figure would be a tetrahedron, the time taken by the ants to converge should be 1/1.5 that is 0.66.

Unfortunately, this solution is wrong. The problem is your assumption that tetrahedral symmetry will be maintained throughout the motion (resulting in constant radial velocity), which is not true given the problem description. As the motion progresses, the tetrahedron will flatten to a square and the corresponding paths will have a total length of $\left(\frac 12 + \frac{\sinh^{-1}(1)}{2\sqrt 2}\right)$ m (approximately 0.81 m).

Answer may be correct but entirely wrong explanation, and time should be 0.725 seconds. How?? We need find the distance ( $S$ ) of centroid from the vertices of tetrahedron and take component of velocity ( $V$ ) along the direction of centorid and find time using fromula $S= V\times T$ . But first we need to find distance of centroid of tetrahedron from vertices $A$ , $C$ , $B$ , $D$ !! To do that we need to find the distance of centroid of triangle $CBD$ (base of tetrahedron) from vertices $C$ , $B$ , $D$ . This follow's normally and distance of centroid of equilateral triangle $CBD$ from vertices $C$ , $B$ , $D$ will be= $\frac { 1 }{ \sqrt { 3 } }$ $m$ . Let this centoid be point $E$ (of triangle $CBD$ ). Now comes the important understanding part!!! We need to find $AE$ on which centroid (of tetrahedron) will lie. The distance $AC$ = $1m$ and $CE$ = $\frac { 1 }{ \sqrt { 3 } }$ $m$ . Therefore using pythagoras theorem $AE$ = $\sqrt { { AC }^{ 2 }- { CE }^{ 2 } }$ $\Longrightarrow$ $AE$ = $\sqrt { 1 - \frac { 1 }{ 3 } }$ $\Longrightarrow$ $AE$ = $\sqrt{ \frac{ 2 }{3} }$ $m$ . Now as we know centroid ( $F$ ) (of tetrahedron) will lie on $AE$ , we need to multiply $AE$ by $\frac{2}{3}$ and will get distance of centroid of tetrahedron $ACBD$ from all the four vertices $A$ , $C$ , $B$ , $D$ . Thus $AF$ = $CF$ = $BF$ = $DF$ = $S$ = $\frac {2 \sqrt { 2 } }{3 \sqrt { 3 } }$ . Difficult part is over and now we will find velocity of approach. Now we are given $V$ = ${ 1 m }/{ s }$ so we need to resolve this $V$ first along $CE$ and from $CE$ to $CF$ . To do this we will multiply $V$ with $\cos 30^\circ$ to get along $CE$ anfd again with $\cos 30^\circ$ to get $V$ along $CF$ . Thus velocity of approach $v$ = $\frac {3}{4}$ ${ m }/{ s }$ . Using formula $S = V \times T$ $\Longrightarrow$ $\frac {2 \sqrt { 2 } }{3 \sqrt { 3 } }$ = $\frac {3}{4} \times T$ $\Longrightarrow$ $T$ = $\frac {8 \sqrt { 2 } }{9 \sqrt { 3 } }$ . Therefore $T= 0.725s$

@Beakal Tiliksew How do you get the velocity of the object in the centroidal direction?I am confused