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course Mth 279
6/28
Query 09 Differential Equations*********************************************
Question: 3.6.4. A 3000 lb car is to be slowed from 220 mph to 50 mph in 4 seconds. Assume a drag force proportional to speed. What is the value of k, and how far will the car travel while being slowed?
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Your solution:
3000v’ = -kv
3000v’ + kv = 0
v’ + kv/3000 = 0
p(t) = k/3000 int= kt/3000
v(t) = Ce^((-kt)/3000)
v(0) = 220
220 = Ce^((-k(0)/3000))
C=220
v(t) = 220e^((-kt)/3000)
50 = 222e^((-4k)/3000)
5/22 = e^(-k/750)
ln(5/22) = -k/750
k = -750ln(5/22) = 1111
how far?
v(t) = 220e^(-1111t/3000)
integrate
x(t) = int (220e^(-1111t/3000)) from 0 to 4
220 * (-3000/1111)e^(-1111t/3000) from 0 to 4
x(4) = 458 miles traveled
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Your procedure is good but the car will not travel anywhere near 458 miles. If you think about the situation, you will realize that this result does not correspond at all.
The problem is in your units. Correct conversion of units, and careful use of units throughout the solution, will yield a more realistic result.
Your method is fine.
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confidence rating #$&*:
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Given Solution:
Correct solution verified by textbook solution
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Self-critique (if necessary):
Why does equation start mv’ = -kv??? I thought the form was mv’ = F -kv. Where is or what is F in this problem???
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Self-critique rating:
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Question: 3.6.6. A vertical projectile of mass m has initial velocity v_0 and drag force of magnitude k v. How long after being fired will it reach its maximum height?
If the projectile has mass .12 grams and after being fires straight upward at 80 meters / second reaches its maximum height after 2.5 seconds, then what is the value of k?
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Your solution:
A. mv’ = -mg - kv
v’ = -g - kv/m
v’ + kv/m = -g
p(t) = k/m q(t) = -g
v = Ce^(-kt/m) + e^(-kt/m) *int(e^(kt/m) *(-g)dt
v = Ce^(-kt/m) + e^(-kt/m)*g(m/k)e^(kt/m)
v = Ce^(-kt/m) - mg/k
v(0) = V_0
v_0 = Ce^(-k(0)/m) - mg/k
v_0 +mg/k = C
v(t) = (v_0 +mg/k)e^(-kt/m) - mg/k
set equal to 0 and solve for t
0 = (v_0 +mg/k)e^(-kt/m) - mg/k
(mg/k)/(v_0 +mg/k) = e^(-kt/m)
ln(mg/k)/(v_0 +mg/k)) = -kt/m
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You're missing a parenthesis:
ln( (mg/k)/(v_0 +mg/k) ) = -kt/m
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t = -m/k * ln(mg/k)/(v_0 +mg/k))
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t = -m/k * ln ( (mg/k)/(v_0 +mg/k) )
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B. find k?
2.5 = -.12/k * ln(((.12*9.8)/k)/(80+ (.12 *9.8)/k))
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You can't solve this equation analytically. You would solve it graphically (plot 2.5 +.12/k * ln(((.12*980)/k)/(8000+ (.12 *980)/k)) vs. k and see where the curve intersects the horizontal axis) or by approximation (e.g., Newton's Method).
A combination of the methods would be most efficient.
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A plot reveals a zero near k = 0.1.
The function is very well-behaved near k = 0.1, so Newton's Method applied to this function, with starting estimate k = .1, would quickly result in an excellent estimate.
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solve for k?
Not sure how to solve for k. Every time I try to solve for k I get held up in the math.
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
Again why is it mv’ = -mg-kv where we used and learned from lectures mv’ = F - kv???
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Self-critique rating:
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Question: 3.6.10. A 82 kg skydiver falls freely for 10 seconds then opens his chute. He reaches the ground 4 seconds later. Assume air resistance is proportional to speed, and assume that with this chute a 90 kg would reach a terminal velocity of 5 m / s.
At what altitude was the parachute opened?
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Your solution:
1) F = mg = (90*9.8) = 880 N
F = kv
kv = mg
k(5) = 880
k = 176 Nm/s
2) mv’ = mg - kv
v’ = g - kv/m
dv/dt = g -kv/m
dv/(g - kv/m) = dt integrate
-m/k * ln|g - kv/m| = t + C
solve for v
ln|g - kv/m| = -kt/m + C
g - kv/m = Ae^(-kt/m)
v = -m/k * Ae^(-kt/m) + mg/k
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Note that the constant A would 'absorb' the constant m / k, so the solution could be written
v = Ae^(-kt/m) + mg/k.
This would give A units of velocity, and would be more intuitive. You could keep the solution in the form you specified, but it would be a little messier to do so.
Generally a constant quantity multiplied by an arbitrary constant will most simply be 'absorbed' by the arbitrary constant.
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I am kind of lost on where to go next to solve for the altitude. At t=10 sec is when the parachute opens. To find the altitude you would have to have position function??
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At this point you want to plug your known quantities in and evaluate the constants A and k.
This gives you a velocity function, which you can integrate to get the position function.
You should obtain a distance between about 60 and 70 meters.
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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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You're doing well. My notes should help with some of the details.
Remember that you're always invited to follow up with a question form.
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