| Author | Maths and Physics (Problems) |
|---|
my bad.. i thought that was in mains paper..x_X
it's from a DPP |
So after a long time..here's a question, maybe just for fun haha..post the answer but not yet the solution please, let everyone try ;)
Q. A particle moves along the positive x-axis, starting form origin (t=0, x=0, velocity = 0). Its acceleration is given by A = 2x^3, where 'x' is it's distance from origin at any point. Find the distance covered by it in 10 seconds. |
| 10km maybe? |
Crazy it may seem
but i think 0 (or negligible) |
but i think 0 (or negligible)
Give method..I think you are getting a 1/infinity or something after integration? Or what is your approach? :)
Gaara has already got it right, just saying :) |
Initial velocity =0
Accn = 0^3=0
So, stationary :3
Anyhow, assuming it made it ahead, you get 1/x on integration.
Hence, tends to 0. |
Vdv/dx= 2x³ so v= -x²
Now dx/dt=-x²
So 1/x =t + C
Here c would tend to infinity so at every t, x would tend to 0 (something went wrong,I have an intuition) |
hmpf
v dv/dx=2x^3 so v=-x^2
now dx/dt=-x^2
so 1/x=t + C
Here c would tend to infinity so at every t, x would tend to 0 (something went wrong,I have an intuition)
for The One Ring:
on origin,the particle is at an unstable equilibrium,so a slight push (a slight movement) would trigger locomotion (however negligible)
for example if acceleration changed with distance as a=x..a slight movement would trigger an exponential escalation in displacement. |
I tried and it gave me -20m lol
I am not cut out for this stuff!! :( |
I did this.
2.2x^3.x = (dx/dt)^2
(-+)2x^2 = dx/dt
(-+)2t = -1/x + C
Now we need something to calculate C. :$ |
| its at infinite displacement from the origin just as there guys stated |
so a slight push
I don't see that anywhere hmmh |
Initial velocity =0
Accn = 0^3=0
So, stationary :3
Anyhow, assuming it made it ahead, you get 1/x on integration.
Hence, tends to 0.
Yea, that's right.
on origin,the particle is at an unstable equilibrium,so a slight push (a slight movement) would trigger locomotion (however negligible)
for example if acceleration changed with distance as a=x..a slight movement would trigger an exponential escalation in displacement.
Where did you assume that "push from"? Besides, what would you define the push to be? Maybe you have a point, I wouldn't be sure of that. |
oh
well i've been taught to assume *a slight push* when these type of question occur mainly by seeing "A particle moves along the positive x-axis" so that the question becomes a "solving" type rather than quickies.
Besides what difference does it make??..even when u put t=infinite in 1/x=t + C the particle is still at rest at (x =0)
Besides, what would you define the push to be
no definition..it's just a simple phrase for causing a slight displacement for further movement |
Besides, what would you define the push to be
no definition..it's just a simple phrase for causing a slight displacement for further movement
Yes I mean what would you say, is it a momentary velociy or momentary acc etc. |
u're taking this too srsly :p
what i meant was the particle got moved to 0.00000000000(and many more 0's)1 somehow from 0 (Just an assumption that the particle made it ahead by some means..still at the end proving it never moved) |
| Stop. |
Stop.
Pun intended. |
| xD |
| integrate root over tan x :p |
