Challenging physics problem | 2D motion

Three particles A, B and C are situated at the vertices of an equilateral triangle ABC of side d at time t = 0. Each of the particles moves with constant speed v. A always has its velocity along AB, B along BC and C along CA. At what time will the particles meet each other?

This is a famous problem from the 2D kinetmatics (Motion in a plane). The best way to attack this, and other such physics numerical, is to change your perspective!

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[chair]

Talking about how so many physics problems can just stump you is really heartwarming. It's nice to know that I'm not the only one who has trouble with encountering a new type of problem.

[parthhooda]

A similar problem was given in the exam for IIT 1984. In it, there was a square instead of equilateral triangle. I read it in a book and in it the solution was given as: (I am giving the solution for the triangle not for square but in a similar way) Firstly, they will meet at a point such that distance from each vertex, the initial positions, is the same. That point will be the centroid. This is because the situation is symmetrical so displacement of all of them will be same. Now the ultimate path taken by partical A (say) will be AO (O is the centroid). The component of velocity of A along this path will be constant and will be equal to vcos(30)=vsqrt3/2. The distance from point A to O will be displacement of A equal to a/sqrt3. The average velocity of A is vsqrt3/2 and displacement is a/sqrt3 so time will be displacement/average velocity= (a/sqrt3)/(vsqrt3/2)= *2a/3v. I found this approach more fun because it only uses basic 9th grade formulas and just a little bit of thinking about what component of velocity will be constant so average velocity along that path will be constant and we will be able to find the time as displacement from initial position to that point/ that component of velocity along that path which remains constant.

[shivamchouhan]

Solved this kind of problems many times using incorrect method during my jee preparation in coaching,
Finally understood the right approach.
Thank you so much

[karthikchowdary]

sir, please dont stop making such videos even if your are getting less views, u are one of the most underrated teachers on yt. If i am not able to understand such concepts it makes me frustated and hate that chapter eventually.But videos like this really help us understand such concepts.Thnaks your very much.

[eshayadav-iycy]

Your method is soo much better than the hcv one, thanks a lottt sir

[varshinilolla]

This video was making me feel regretful that I hadn't even tried it, and at the same time realize how remarkably elegant the solution could be no matter how tricky the question might seem!
Love your videos!!!
P.S. This is question from Problems on general physics by IE Irodov

[sujalpatel]

I was so confused in this Question
Thank you so much for better understanding ☺️

[aryanraj]

aha, i guessed the same problem!
And as always preety explanation with real life philosophy.

[vaidehirathore]

Loved this one... I learnt a new perspective of viewing relative motion. Thanks sir☺

[Raian-oxdb]

I tried something similar, instead of the length = 0, I thought when the 3 particles meet the area of the triangle should be 0. Thank you so much for the wonderful presentation.

[peacesh]

Great I enjoyed the new perspective .... .... Great work

[Yguy]

I've watched a lot of videos from you In khanacademy. Nice explanation!

[medianinitiantion]

This question was in Concept of physics by Hc verma. It was in example but there was another question relating to hexagon instead of a triangle

[Catastatic]

the answer to the first problem is time = a*Pi/(V*sqrt(3)). Each person moves along a logarithmic spiral with length = a*Pi/sqrt(3)

[prayagpr]

6:57 If there is a component along the triangle towards A then there is also a component perpendicular to it wouldn't that also effect the triangle?

7:53 And also sir, is it necessary that the time taken to meet A and B will be the same for all 3 of them to meet simultaneously?

[AnweshSarthak]

The motion is totally understandable for those who have had watched beyblade

[palanibharathi]

Thank you a lot for this wonderful explanation...

[brolovesanime]

i am in class 9th and i tried to attempt this and got t = a/3v and this is because i took sec theta instead of cos cannot believe i came this close man thank you so much...

[unifactsbyparth]

Abj sir's class illustrations in relative motion

[themidnight-man]

Ohh
I was just solving this question yesterday in HCV...