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course Mth 279
6/30 10:42 pm, partial submission, I need to spend more time on this section.
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:
mv dv/dx = -kv/(1+x)
vdv = -kv/(m (1+ x)) dx
1/2 v^2 = -kv/m ln (1+x) + C
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If you divide both sides of the equation
vdv = -kv/(m (1+ x)) dx
by v you get
dv = -k/(m (1+ x)) dx,
which does not lead 1/2 v^2 on the left-hand side, nor will v be a factor of the right-hand side or its integral.
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v = sqrt(-2kv/m) * 1/2 ln(1+x) + C
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This does not follow from the preceding.
The solution, found by multiplying both sides by 2 and taking the square root of each side, would be
v = +-sqrt( -2 kv/m ln(1+x) + C)
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However the preceding didn't follow correctly from the original equation.
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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:
m dv/dt = -mg + kv^2 ( assuming v_0 is positive)
dv/dt = -g + k/m v^2
dv - k/m v^2 dv = -g dt
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There is no dv on the k/mv^2 term.
You can separate this equation, obtaining
dv / (k/m v^2 - g) = dt
The left-hand side is integrated using partial fractions, noting that (a - b) factors as (sqrt(a) + sqrt(b)) ( sqrt(a) - sqrt(b)). Be sure to multiply this expression out using the distributive law, in order to see why it is so.
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v - (1/3) k/m v^3 = -gt + v_0
max height occurs when v = 0
0 = -gt+ v_0
I have too many variables. ???
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Your solution will involve two constants, k and your integration constant.
The condition on the initial velocity gives you one equation.
The maximum height gives you another.
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I stopped here because I keep running into the problem of not being able to solve for k. Hopefully feedback on the first two will help me solve the others.
confidence rating #$&*:
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Question:
3.7.8. Mass m accelerates from velocity v_1 to velocity v_2, while constant power is exerted by the net force. At any instant power = force * velocity (physics explanation: power = dw / dt, the rate at which work is being done; w = F * dx so dw/dt = F * dx/dt = F * v).
How far does the mass travel as it accelerates?
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Your solution:
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Hint:
P = F_net * v,
where P is power and F_net is net force.
F_net = m v '
by Newton's Second Law.
These two relationships can be combined to obtain a differential equation in v, which is not difficult to solve.
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Given Solution:
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Question:
3.7.11. An object falls to the surface of the Earth from a great height h, and as it falls experiences a drag force proportional to the square of its velocity. Assume that the gravitational force is - G M m / r^2.
What will be its impact velocity?
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Your solution:
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Hint:
The net force is m v ' = m v dv/dr, where r is the position measured relative to the center of the Earth.
The net force is the sum of the gravitational force and the drag force.
You are given the gravitational force.
Assume drag force with constant k.
What is your equation, and what is its general solution?
How do the specific conditions of this problem fit with your general solution?
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confidence rating #$&*:
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Given Solution:
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Question:
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Question:
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#*&!
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Check my notes and see if you can make some more progress.
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