hey this is a question i have for Year 12 Maths C in Australia. It is irritating me no end. <line of text deleted by CaptainBlack>
here we go...
A particle of mass one unit moves in a straight line in the horizontal plane against a resistance v+av^2 , where a is a positive constant. If v is the velocity of the particle at time t , then dv/dt=-(v+av^2) represents the particle’s motion.
Given this particle is released from the origin, O, with an initial velocity of u, find the velocity of the particle v as a function of the time t .
As the value of t increases, what happens to the velocity of this particle? Would this value change if a is negative?
PLEASE HELP! I WILL BE FOREVER IN YOUR DEBT. EMAIL ME OR RESPOND HERE, I WILL CHECK IT EVERY 30 MINUTES
THANKS
If the initial velocity is positive the acceleration is negative and so the particle slows down and as .
If the initial velocity is negative, the velocity will change until the acceleration is zero, that is , that is or . The first of these will not occur as all trajectories starting with negative velocity are repelled by this solution. The second is an attractor, that is all trajectories of which start negative more towards it.
RonL
Hello,
CaptainBlack wrote:"The first integral is done by resolving the integrand into partial fractions, and the second is -t."
I'm going to show you what he meant:
. I assume that the integral of the right side isn't a problem to you.
Take . Thus you know that
(There isn't a summand in the numerator of the fraction which contains the variable v). Solve for N and M:
N = 1 and M = -a. Therefore your integral becomes:
1.
2. Use integration by substitution here: u = 1+av; u' = a
3.
Without considering the bounds you have now:
. Now de-logarithmize(?) using the base :
. Multiply both sides by the denominator:
I'll leave the rest for you