1(a) is correct....for sure.....
A particle is restricted to move on x-axis. The particle must be moving towards origin if,
A)x\frac{dx}{dt}}<0 B)x\frac{d^{2}x}{dt^{2}}<0 C) (\frac{dx}{dt})(\frac{d^{2}x}{dt^{2}})<0 D)x^{2}\frac{dx}{dt}<0
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17 Answers
11I think
C)should be the answer
Since it is
V*A<0
V is velocity and A is acceleration.
V and A will be antiparallel therefore the product will always be less than 0
1No..C cannot be the answer
Consider a situation in which the particle is moving to the right(pos. X) and IS to the right of the origin. Now consider it a car. So if u apply the brakes, U will now have negative acc. but still with pos. velocity..coz ur still moving in the postiive direction. Therefore, the product in C, would be negative..but the particle would NOT be moving towards the origin. So, A is the only one which satisfies any scenario
1No i doubt that B will be an answer coz the acceleration could be negative, but the particle could still be moving towards + X axis because neg. acc. could just mean velocity in forward direction is decreasing, without the particle necesarily having to turn direction and start to go toward the origin
49@virang i fear that i feel that u r wrong in the sense how does V.A<0 proves that particle is moving towards origin?????this is not shm.....[7][7][7][7]
49[3][3]
varied thoughts.........i hope some one will come up with the last option too...[3][3]
1ans=D
it can be riten as dx/dt<0..(as x2 is +ve).
wich means its postion is decreasing wrtt time..ie moving towards (0,0)
1Anirudh is right
Flaw in Avinau's explanation is, the particle could be moving to the left..but beyond the origing...freeze an instant of time where particle is at x<0. It would still be moving to the left, but no longer towards the origin
Anirudh's answer satisfies that regardless of whether the particle is to the right or left of origin, it will still be oving towards the origin
1A) should be correct .
if x>0 then V<0
and if x<0 then V>0
thus particle moves towards origin at any x
1Thank u for the ans.
I exactly don't know the correct ans.
Ur ans. seems to be correct though.