Continuos Systems?
Consider the vertical motion of a particle P. Let t denote time and v(t) the vertical velocity of the
particle measured so that a positive v is directed upwards. The acceleration of P is
dv/dt = −g − kv|v|
where g is the constant (and positive) acceleration due to gravity and k is a positive constant. In this
model, the air resistance is proportional to the square of the velocity and the modulus sign ensures
that the resistance always opposes the velocity.
Use a phase-plane diagram to investigate the behaviour of P for each of the following initial conditions.
(a) v(0) = 0, (b) v(0) =sqrt(g/k) , (c) v(0) = -2*sqrt(g/k)
For each case, describe in a few words the behaviour of the displacement, velocity and speed during
the subsequent motion of the particle. Assume that the particle is sufficiently high for the ground to
be unimportant.
(Hint: When sketching the phase plane apply the definition that |v| = v for v 0 and |v| = −v for
v < 0.)
Here I can’t draw any phase plane diagrams, but I will just describe the general physics of each scenario:
(a) the particle P is simply dropped and will accelerate under gravity g, but air resistance creates a drag force on the particle and will slow the particle’s descend. Eventually the particle will reach terminal velocity.
Terminal velocity is reached when dv/dt = 0
So g = kv² or v = sqrt(g/k)
(b) the particle P is shot up vertically with initial velocity v(0) = the terminal velocity sqrt(g/k). Eventually the particle under deceleration from gravity and air resistance will slow and then stop, and then start to fall downward towards the ground and eventually reach a downward speed equal to the termial velocity.
(c) Here the particle P is shot downwards at twice the speed of the termial velocity. Here what happens depends on the ratio of g/k, or in other words, how strong is the air resistance relative to the gravitational force.
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