I've even got a starter question to get you guys into the scenario.
Once you've completed the starter question, under the solution comment attaches the main question, which is unsolved.
this post was submitted on 17 May 2024
9 points (100.0% liked)
Daily Maths Challenges
189 readers
1 users here now
Share your cool maths problems.
Complete a challenge:
- Post your solution in comments, if it is exactly the same as OP's solution, let us know.
- Have fun.
Post a challenge:
- Doesn't have to be original, as long as it is not a duplicate.
- Challenges not riddles, if the post is longer than 3 paragraphs, reconsider yourself.
- Optionally include solution in comments, let it be clear this is not a homework help forums.
- Tag [unsolved] if you don't have a solution yet.
- Please include images, if your question includes complex symbols, attach a render of the maths.
Feel free to contribute to a series by DMing the OP, or start your own challenge series.
founded 6 months ago
MODERATORS
you are viewing a single comment's thread
view the rest of the comments
view the rest of the comments
Kind-of solution
F = ma and m = 1, so a = F = 1/r^2. I used a = -1/r^2 so the force experienced by the particle at r = 5 would be negative. a = dv/dt.dv/dt = -1/r^2
v dv/dt = -1/r^2 dr/dt → v = dr/dt. Multiply both sides by v, but write it as dr/dt on the right
v dv = -1/r^2 dr → Multiply both sides by dt
0.5v^2 = 1/r + C → Integrate
v^2 = 2/r + C → Multiply both sides by 2, find C = -2/5 by plugging in r = 5, v = 0
v^2 = 2/r - 2/5
5*arcsin(sqrt(r/5)) - sqrt(5r-r^2) = sqrt(2/5) t + C → Solve the differential equation y'^2 = 2/y - 2/5 with Wolfram Alpha, find C = 5π/2 by plugging in r = 5, t = 0
5*arcsin(sqrt(r/5)) - sqrt(5r-r^2) = sqrt(2/5) t + 5π/2 → Can find that the particle reaches the origin at t = 5sqrt(5)π / 2sqrt(2) ≈ 12.418 seconds
I made heavy use of internet resources on this one, not just for solving the differential equation but also for technique - I got a bunch of wrong answers before getting this one that I finally think is right, at least up until the moment of origin contact
Graph (the hidden ellipse is because I thought the path may be elliptical, but it doesn't appear to be)