okay now this is a very simple question, yet i feel like killing someone because im not getting the answer out.
Q: A particle moves in a straight line away from a fixed point O in the line, such that at time t its displacement from O is x and its velocity is v. At time t=0, x=0, v=V. Subsequently the particle is slowing down at a rate proportional to the square of its speed. Find an expression for the velocity v in terms of displacement x.
What i did was v.dv/dx = -kv^2
and then work from there... i ended up with v = e^-kx + V
PLEASE HELP ME AS IM NOT SURE WHAT I DID WRONG!! D:
Thanks
Q: A particle moves in a straight line away from a fixed point O in the line, such that at time t its displacement from O is x and its velocity is v. At time t=0, x=0, v=V. Subsequently the particle is slowing down at a rate proportional to the square of its speed. Find an expression for the velocity v in terms of displacement x.
What i did was v.dv/dx = -kv^2
and then work from there... i ended up with v = e^-kx + V
PLEASE HELP ME AS IM NOT SURE WHAT I DID WRONG!! D:
Thanks
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