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 |