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course MTH-279
06/25 around 11:45AM.
Query 10 Differential Equations*********************************************
Question: 3.7.4. Solve the equation m dv/dt = - k v / (1 + x), where x is position as a function of t and v is velocity as a function of t.
How far does the object travel before coming to rest?
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Your solution:
The Equation, has the variables (v,x,t)
m dv/dt = - k v / (1 + x)
Using dv/dt = (dv/dx)(dx/dt) = (dv/dx)v
Equation becomes: (with only variables v,x)
mv dv/dx = - k v / (1 + x)
dv/dx = -k/(m(1+x))
Integration of both sides yeilds:
v(x) = (-k/m)*ln(1+x) + c
v(0) = c = v_0 due to ln(1) = 0
letting v = 0, and solving for x:
0 = (-k/m)*ln(1+x) + v_0
v_0 = (k/m)*ln(1+x)
(m/k) v_0 = ln(1+x)
1 + x = e^(mv_0/k)
x_f = e^(mv_0/k) - 1
which is the final position of x at rest.
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Good.
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confidence rating #$&*:
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Given Solution:
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Question:
3.7.6. A vertical projectile has initial velocity v_0, and experiences drag force k v^2 in the direction opposite its motion. Assume that acceleration g of gravity is constant. Find the maximum height to which the projectile rises.
If the initial velocity is 80 m/s and the projectile, whose mass is .12 grams, rises to a height of 40 meters (estimated quantities for a plastic BB), what is the value of k?
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Your solution:
mv' = F - kv^2
v' = F/m - kv^2/m
v'/(F/m - kv^2/m) = 1
dv/(F/m - kv^2/m) = dt
dv/((F/m)(1 - kv^2/F)) = dt
dv/(1 - kv^2/F) = (F/m) dt
dv/((1 - kv/F)(1 + kv/F)) = (F/m) dt
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Factoring 1 = k v^2 / F you would get (1 - v sqrt(k/F)) (1 + v sqrt(k/F) )
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The differential equation better solved in terms of v and x.
dv/dt = dv/dx * dx/dt = dv/dx * v
so the equation becomes
m v dv/dx = -k v^2 - m g.
To see if you're on the right track, an intermediate step of the integration is
ln | k / m v^2 + g | = -2 k / m * x + c
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Integration of both sides yeilds:
(1/2)(sqrt(k/F))*ln((1+sqrt(k/F))/(1-sqrt(k/F))) = (F/m)t + c
(1+sqrt(k/F))/(1-sqrt(k/F)) = e^(2sqrt(Fk)t/m + c) = Ce^(2sqrt(Fk)t/m)
(1+av)/(1-av) = B
v = (1/a)((B-1)/(B+1))
v = sqrt(F/k)((Ce^(2sqrt(Fk)t/m) - 1)/(Ce^(2sqrt(Fk)t/m) + 1))
Divide by e^("""" """")
v = sqrt(F/k)((C - e^(2sqrt(Fk)t/m))/(C + e^(2sqrt(Fk)t/m)))
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mv' = mg - kv^2
Using the previous equation, F = mg:
v = sqrt(mg/k)((C - e^(2sqrt(mgk)t/m))/(C + e^(2sqrt(mgk)t/m)))
Max height would be at v = 0
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v_0 = 80
m = 0.12E-3
If v(x), v(40) = 0 is max height
k?
at t = 0
v(0) = sqrt(mg/k) (c/c) = 80
80 = sqrt(mg/k) = sqrt((0.12E-3)(9.81)/k)
k = (0.12E-3)(9.81)/80^2 = 1.84E-7
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Question:
3.7.8. Mass m accelerates from velocity v_1 to velocity v_2, while constant power is exerted by the net force. At any instant power = force * velocity (physics explanation: power = dw / dt, the rate at which work is being done; w = F * dx so dw/dt = F * dx/dt = F * v).
How far does the mass travel as it accelerates?
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Your solution:
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F_net = P/v, where P is the constant power.
You will want to solve an equation in terms of v and x (if necessary see previous note on this change of variable).
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Given Solution:
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Question:
3.7.11. An object falls to the surface of the Earth from a great height h, and as it falls experiences a drag force proportional to the square of its velocity. Assume that the gravitational force is - G M m / r^2.
What will be its impact velocity?
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Your solution:
F_net = kv^2 - GMm/x^2
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Again you'll want to express this in terms of v and r, so that m dv/dt becomes m v dv/dr/
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
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See if you can make some more progress on these problems in terms of my notes.
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