#$&*
course MTH 279
2/14 330 pm
Query 09 Differential Equations
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Question: 3.6.4. A 3000 lb car is to be slowed from 220 mph to 50 mph in 4 seconds. Assume a drag force proportional to speed. What is the value of k, and how far will the car travel while being slowed?
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Your solution:
k = 4,000,000 (approx) or 270,000*ln(50/220) (exact) I changed seconds into hours instead of changing mph into mps
m(dv/dt) = -kv I solved this as a linear first order homogeneous equation and got:
v=v_0e^(-k/m t)
v(1/900) = 50 = 220e^(-k/3000 * 1/900) and found k.
x = .1275 mi
using the function: x=(mv_0 / k) * (1-e^(-k/m t)
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This looks good, but you don't mention your value of k, and your result doesn't agree with the result I obtained (.1275 miles is about 600 feet), whereas my solution shows a distance of about 280 feet).
I don't see any flaws in your work, except that the mass of the car is not 3000 lb, since 3000 lb is a force rather than a mass.
You can solve
v(1/900) = 50 = 220e^(-k/m * 1/900)
for k/m.
The mass of the car is
mass = weight / acceleration of gravity = 3000 lb / (32 ft/s^2) approximately, so that
k = (k/m) * 3000 lb / (32 ft/s^2).
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary): OK
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Self-critique rating: OK
@&
Compare your results to mine:
My solution, starting with
v = A e^(-k /m * t) , A > 0,
which is the same as your solution except that I'm using A instead of v_0, is as follows:
v(0) = 323 ft / s
v(4 s) = 73 ft / s
so
A e^0 = 323 ft / s, giving us A = 323 ft / sec.
A e^(-k / m * 4 s) = 73 ft / s
so
e^(-k / m * 4 s) = 73/323 = .28, very approximately, and
-k / m * 4 s = ln( . 28 )
k / m = -ln(.28) / (4 s) = .31 s^-1.
The distance the car will travel is
integral ( v(t) dt, t from 0 to infinity)
= integral( 323 ft / s e^(-.31 s^-1 * t), t from 0 to infinity)
= -1/.31 s * 323 ft / s * (e^(-.31 s^-1 * 4 s) - 0)
= 1000 ft * .28 = 280 ft, very approximately.
The value of k, which was also requested is
k = (k/m) * m = .31 s^-1 * m,
with m = weight / accel of gravity = 3000 lb / (32 ft/s^2) = .94 lb s^2 / ft, so that
k = .31 s^-1 * .94 lb s^2 / ft = .29 lb s / ft.
.94 lb s^2 / ft is .94 slug so the value of k could also be expressed as
k = .29 slug / s.
When multiplied by v to get the magnitude of the drag force, this would yield units of slug / s * ft/s = slug * ft/s^2. The slug being the unit of mass in the ft-lb-second system, and the ft / s^2 the unit of acceleraiton, the slug ft / s^2 is the unit of force, more commonly called the pound.
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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:
t= (-m/k)ln( gm / [k(v_0 + gm/k)]
starting equation: mv' = mg - kv solved as a first order linear non homogeneous with initial condition v(0) = v_0 to get:
v = - gm/k + (v_0 +gm/k)e^(-k/m t)
max height would occur when v=0, so plugged in 0 for v and solved for t.
For the second part, I'm not sure how would could find k with the given information. Looking at the original equation
mv' = mg - kv
if v=0 as it is at the time specified, then the only term with k in it disappears.
@&
mv’ = -mg-kv
v’ = -g-kv/m
v’ + kv/m = -g
Solving as a linear nonhomogeneous equation, we find integrating factor
e^(integral(k/m dt)) = e^(kt/m)
Our equation becomes
(e^(kt / m) v) ' = -g e^(kt / m)
Integrating both sides we get
e^(kt / m) v = integral (- g e^(kt/m) dt)
e^(kt / m) v = -m g / k e^(kt/m) + c
so that
v = -m g / k + c e^(-kt/m)
Initial velocity v_0 at t = 0 implies that
v_0 = -m g / k + c
so that c = v_0 + m g / k
and our solution is
v = -m g / k + (v_0 + m g / k) e^(-k t / m).
To find the maximum height first let v(t) = 0 and solve for t:
-m g / k + (v_0 + m g / k) e^(-k t / m) = 0
e^(-k t / m) = ( - m g / k ) / (v_0 + m g / k)
so that
t = -m/k * ln(( mg/k )/ (v_0 + mg/k) )
Using the values given:
2.5 s = -0.00012 kg / k * ln((0.00012 kg * 9.8 m/s^2 / k) / (80 m/s + 0.00012 kg * 9.8 m/s^2/ k))
Letting C stand for .00012 kg / k, we will write the equation in terms of C. Having found, or approximated, the value of C we can then divide .00012 kg by C in order to get k.
We will also express the acceleration of gravity as g, knowing that we can replace this symbol with 9.8 m/s^2 at any time.
The equation becomes
2.5 s = -C ln (C g / (80 m/s + C g) )
Since ln (C g / (80 m/s + C g) ) is unitlss, the unit of C must be seconds, in order to make the units of the left- and right-hand sides identical.
C g will then have units of s * m/s^2 = m/s, which makes the units of the numerator and denominator of the expression C g / (80 m/s + C g) , and confirms the unitless quality of the natural log.
Knowing that the units are consistent throughout, we can treat the equation numerically as
2.5 = -C ln( C * 9.8 / (80 + C * 9.8) )
Since C occurs both within and outside of the natural log expession, it is impossible to solve exactly for C.
Plotting the expression
-C ln( C * 9.8 / (80 + C * 9.8) )
or using trial and error we find that the value 2.5 is attained when C = 1.24, approximately.
So C = 1.24 seconds, and k = .00012 kg / (1.24 s) = .0001 kg / s.
In this case the velocity function will be
v(t) = -m g / k + (80 m/s + m g / k) e^(-k/m t)
m g / k = .00012 kg * 9.8m/s^2 / (.0001 kg / s) = 12 m/s, approx..
k/m = ( .0001 kg / s ) / (.00012 kg) = .83 s^-1, approx.
So
v(t) = -12 m/s + (80 m/s + 12 m/s) e^(-.83 s^-1 t)
v(t) = -12 m/s + 92 m/s e^(-.83 s^-1 t).
A plot of this function shows curve, concave up, through (0, 80) and (2.5, 0).
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary): I'm confident in the first part of my answer, but I don't know how to determine k from the given information.
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Self-critique rating: 3
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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:
0.7 meters (that can't be right)
With the simplifying assumption that g is constant:
mv' = -mg + kv that is, air resistance is opposing the force due to gravity in this scenario after the parachute is opened.
v = mg/k + v_terminal * e^(k/m t)
y = (mg/k)t + m*v_terminal/k(e^(k/m t) -1)
making the substitution that k = -mg/v_terminal
y= -t/v_terminal -(1/g)( e^(g/v_terminal t) - 1)
using t = 4, that is, the change in time between which the parachute is deployed and the person hits the ground, leads to the above solution.
@&
Good. If you plug the given values into the equation you should get something that agrees with the following:
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@&
The terminal velocity of the skydiver is m g / k, so after the chute is opened the value of k is found by solving the equation
v_term = m g / k
for k, obtaining
k = m g / v_term = 90 kg * 9.8m/s^2 / (5 m/s) = 180 kg / s.
Assuming that the skydiver's velocity when opening the chute is the -98 m/s that would result from free fall with no drag (this is not a valid assumption, and opening the chute at this speed would be very dangerous, but no other information was provided in the problem) we use the solution
v(t) = mg/k + (v_0 - mg/k)e^(-kt/m)
to the equation m v ' = -k v + m g.
m g / k will be 90 kg * 9.8 m/s^2 / (180 kg/s) = 5 m/s, the terminal velocity (as we already know).
The velocity function becomes
v(t) = 5 m/s + (98 m/s - 5 m/s) * e^(-180 kg/s * t / (90 kg)) = 5 m/s + (93 m/s) e^(-2 s^-1 t).
The distance fallen during 4 seconds will be the integral of v(t), with t ranging from 0 to 4 seconds:
integral ( (5 m/s + (93 m/s e^(-2 s^-1 t) ) dt, t from 0 to 4 s)
is calculated as the change in the antiderivative function
5 m/s * t - 46.5 m * e^ (-2 s^-1 t)
from t = 0 to t = 4.
At t = 0 the antiderivative function has value -46.5 m.
At t = 4 s the antiderivative functin has value 20 m - .02 m = 19.8 m.
So the distance fallen is the integral, with its value 19.8 m - (-46.5 m) = 66 m, approx..
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary): I think I modeled my equations correctly, but my error must have been in interpreting the significance of v_terminal or maybe my value for k. In looking at the equation for v, if you let time approach infinity, then the velocity also approaches infinity. Maybe I should've left the force equation as -mg -kv?? In which case I modeled my equations incorrectly. But I modeled it in that way to demonstrate that drag was opposing the force of gravity, which I think is a correct statement.
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The equation is
F_net = -k v + m g
where we regard the downward direction as positive.
This becomes
m v ' = - k v + m g,
and with initial condition v(0) = v_0 leads to the solution
v(t) = mg/k + (v_0 - mg/k)e^(-kt/m)
Note the negative coefficient of t in the exponent.
*@
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Self-critique rating: 3"
&#This looks good. See my notes. Let me know if you have any questions. &#