Four Ants - HC Verma

Solved a more difficult problem in HC Verma 5-6 years back.. :)

Four bugs are on the corners of a 1 meter square. Each bug always faces the next bug (on the next clockwise corner). If they all walk forward at the same speed until they meet, how far does each bug travel?


Solution: 
Highlight the part between the * symbols for the answer.
*Since the bugs always move perpendicular to each other, so when they meet at the center they have traveled a distance equivalent to their initial separation i.e. 1 m.

If some of you remember, in the HC Verma problem, the three ants were moving in a equilateral triangle of side-size 1. The three ants were such that 1 is following 2, 2 is following three and three is following 1. By symmetry, they would meet at the centre. The time taken was 2/3v as you had to take components along the line joining the two ants. So, v(1+cos60) was the real relative speed along the line joining two ants. So, time taken was 2/(3v). So, the distance traveled was 2/3. :)*

Comments

  1. Unknown6:41 PM, November 06, 2009

    what is the solution? is it really 1!!!

    the answer I got is :
    1- distance travelled by each ant is Pi/4
    2- Straight line distance from A to B is 1/2

    ReplyDelete
  2. Pratik Poddar8:28 PM, November 06, 2009

    The answer is 1. :)

    We want the distance traveled by each ant. Solution goes as follows.

    The ants keep constant speed throughout their motion. Their trajectory is extremely complex. So, geometrical solution will not help. We calculate the time taken for the ants to meet. Now, we see that ant 1 s always moving towards ant 2 and ant 2 towards ant 3. They initially form a square. By symmetry, they always form a square. So, ant 1 is always moving perpendicular to the line joining ant 2 and ant 3. This means, at all times, the velocity of ant 1 is perpendicular to that of ant 2. So, the relative velocity of ant 1 towards ant 2 is always v. So, the time taken to make this distance zero is 1/v.

    So, the distance traveled total is 1 m. :)

    Hope that helps!!

    ReplyDelete
  3. suyash12:35 PM, November 08, 2009

    i remember having solved the HC Verma problem myself, and perhaps would have managed this one too (considering its simpler) but feeling too lazy to exercise my brains, i saw the solution. nice.

    ReplyDelete
  4. Unknown5:25 PM, September 29, 2012

    by same arguments relative velocity of ant 1 with respect to ant 3 is 2v and pointing towards 3 .
    in fact trajectory of ant 1 in reference frame of ant 3 is straight line with speed 2v..
    so time taken to meet is (square root 2)/2v. hence answer is (square root 2)/2..i don't know which one is right .

    ReplyDelete
  5. Unknown5:30 PM, September 29, 2012

    in the reference frame of ant 3 ,ant 1's trajectory is straight line with speed 2v.
    (because velocity of ant1 is perpendicular to ant2 velocty and velocity of ant2 is perpendicular to ant 3 velocty so ant 1 velocity and ant 3 velocity are in opposite direction. )
    hence the answer is square root(2)/2.
    i don't know why the answers are not matching..please do sort out this issue .

    ReplyDelete
  6. Pratik Poddar12:58 PM, September 08, 2026

    Ha, fair enough. The 4-ant square version is indeed the kind one - ant 2 always moves perpendicular to ant 1's line of sight, so it contributes nothing to the closing speed, and each ant just walks 1 m. (replied using AI)

    ReplyDelete
  7. Pratik Poddar12:59 PM, September 08, 2026

    The answer 1 m is right; here is where your argument slips. It is true that v1 = -v3 at every instant (180-degree rotational symmetry), so the relative velocity of ant 1 in ant 3's (translating) frame has magnitude 2v. But that relative velocity is NOT directed along the line joining ant 1 to ant 3 - v1 points along the side of the (shrinking, rotating) square, making 45 degrees with the diagonal. Only the component along the diagonal closes the gap: 2v*cos(45) = sqrt(2)*v. Initial diagonal separation is sqrt(2) m, so time = sqrt(2)/(sqrt(2) v) = 1/v, and distance walked = v * (1/v) = 1 m. Consistent. Also, since the relative velocity rotates with time, the trajectory in ant 3's frame is not a straight line. (replied using AI)

    ReplyDelete
  8. Pratik Poddar1:01 PM, September 08, 2026

    Answered in my reply to your other comment - short version: the relative velocity has magnitude 2v but rotates; it is not locked onto the line joining ants 1 and 3, so the straight-line-at-2v picture is the flaw. The gap closes at sqrt(2)*v, giving the same 1 m answer. (replied using AI)

    ReplyDelete

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