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course Mth 279
4/8
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:
mv' = F_net = -kv
v' = -(k/m)v
dv/v = -(k/M)
Int(dv/v) = -(k/M) int(dt)
ln(v) = -kt/m + c
v = e^(-kt/m + c) = Ce^(-kt/m)
v(0) = 220
v(0) =Ce^0 = C = 220
v(t) = 220e^(-kt/m)
v(4) = 50
50 = 220e^(-kt/m)
5/22 = e^(-kt/m)
Solving the equation for k:
(-ln(5/22)m)/4 = k
(-ln(5/22)(3000/32.2)/4 = k
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k has units, as do the other quantities (4 is 4 seconds, 3000 is 3000 lbs, 32.2 is 32.2 ft/sec)
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k = -34.5
which makes the equation:
v(t) = 220e^(-0.37t)
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Good, but this would be better related to the conditions by including units:
v(t) = 220 ft/sec * e^(-.037 s^-1 t)
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and since v(t) = dx/dt, and x = x(t)
Int (from 0 to 4) of 220e^(-0.37t)
= 220 (Int (from 0 to 4) of e^(-0.37t))
= 220 ((-2.7e^(-0.37(4))) - (-2.7e^(-0.37(0))))
= 220 (2.09) = 459.38 miles
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You aren't likely to end up with the right units if you don't use them throughout.
If you use miles/hour as your constant and integrate with t in seconds, you will get a result in miles/hour * seconds.
220 miles / hour * 2.09 seconds is not 459.38 miles.
220 miles / hour * 2.09 seconds is about 670 feet.
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Think about what happens during a typical increment `dt of time containing sample clock time t_sample. The speed is 220 mph * e^(-.37 s^-1 t_sample) and the duration of the interval is `dt. The sample speed fo the interval is 220 mph * e^(-.37 s^-1 * t_sample), the duration of the interval is `dt, where t_sample and `dt are in the same units. Since .37 s^-1 is in units of s^-1, t_sample and `dt will be most conveniently expressed in seconds.
Multiplying these two quantities we get the distance traveled during the interval.
However the unit mph multiplied by the unit second doesn't work out that conveniently. Better to express mph in miles / second, or feet / second, or kilometers / second or cm/second, or anything with second in the denominator, so you don't need to do unit conversions in your final calculation.
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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 reaches its maximum height after 2.5 seconds, then what is the value of k?
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Your solution:
mv' = F_net
mv' = -kv - mg
v' = -(k/m)v - g
dv/dt = (-kv/m - g)1
dv/(kv/m + g) = -1
Integration of both sides
Left side:
u = kv/m + g
du = k/m dt
dv = m/k du
(m/k)(ln(u)) = -t + c
ln(u) = (-t + c)(k/m)
u = e^((-t + c)(k/m))
= Ce^(-kt/m)
kv/m + g = Ce^(-kt/m)
kv/m = Ce^(-kt/m) -g
v = ((Ce^(-kt/m) -g)m)/ k
and since c and k are constants
v(t) = Ce^(-kt/m) - mg/k
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Max height at v(t) = 0
v(t) = Ce^(-kt/m) - mg/k
at v(0), C = v_0 + mg/k
Ce^(-kt/m) = mg/k
-kt/m = ln((mg)/(kC))
t = (-(ln((mg)/(kC)))m)/ k
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Given values:
Mass = 0.12g
v(0) = 80 m/s
C - mg/k = 80
t = (-(ln((mg)/(kC)))m)/ k
2.5 = (-(ln((0.12*10^-3(9.81))/(kC)))(0.12*10^-3))/ k
My writing stoped making sence after that
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Your function is
v(t) = Ce^(-kt/m) - mg/k
with C = v_0 + m g / k, so
v(t) = (v_0 + m g / k) e^(-k/m * t) - m g / k.
v(t) = 0 when
(v_0 + m g / k) e^(-k/m * t) - m g / k = 0
so that
e^(-k/m * t) = (m g / k) / (v_0 + m g / k)
with solution
t = -m/k * ln ( (m g / k) / (v_0 + m g / k) ).
It's not possible to get a closed-form solution for k, but for the given values of m and v_0, with g = 9.8 m/s^2, the expression on the right becomes a function of k. It is possible to graph this function and by one method or another determine the value of k that yields t = 2.5 seconds.
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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:
90 kg (man + chute) Terminal Velocity = 5 m/s
V_T = -F/k = -mg/k = ((90)(9.81))/ k = 5
k = 176.58
Altitude?
mv' = F_net = kv + mg
Counted down as positive
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If down is positive then the drag force is upward, so your equation should be
m v ' = -k v + m g.
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dv/dt = kv/m + g
Int (dv/(kv/m + g)) = Int (dt)
u = kv/m + g
du = k/m dt -> dv = m/k du
(m/k)(ln(u)) = t + c
ln(u) = (t + c)(k/m)
u = e^((t + c)(k/m))
= Ce^(-kt/m)
kv/m + g = Ce^(-kt/m)
kv/m = Ce^(-kt/m) -g
v(t) = ((Ce^(-kt/m) -g)m)/k
and since C,k are constant
v(t) = Ce^(kt/m) - mg/k
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v(0) = ?
82kg falls for 10s (v_f = v_i + at)
v_f = (82)(10) = 820
v(0) = 820
820 = Ce^0 - mg/k
820 = C - ((82)(9.81))/(176.58)
C = 825
And since v(t) = dx/dt, x = x(t)
Int (from 0 t0 4) 0f 825e^((176.58/82)t)
383(e^2.15t), t =from 0 to 4
383(e^8.6 - e^0) = 2080
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After 10 seconds of free fall the skydiver would reach speed v = 9.8 m/s^2 * 10 s = 98 m/s, over 200 mph. This is probably unrealistic, since the terminal velocity of a tumbling human body is only about 120 mph. However by minimizing air drag it is possible to exceed this speed, though I'm not sure about opening a parachute at this speed. (OK, just checked an Internet source and found the following quote: "However, by diving or "standing up"in free fall, any experienced skydiver can learn to reach speeds of over 160-180MPH. Speeds of over 200MPH require significant practice to achieve. The record free fall speed, done without any special equipment, is 321MPH. Obviously, it is desirable to slow back down to 110MPH before parachute opening.". So apparently 98 m/s, which is around 230 mph, is possible, opening a parachute at this speed isn't something you would want to try at home).
Terminal velocity for mass 90 kg is 5 m/s^2, so at that speed
F_gravity = 90 kg * 9.8 m/s^2 = 880 N, approx.
F_resistance = k v
and the two are equal, so
k * 5 m/s = 880 N
k = 176 N s / m.
The equation of motion after the parachute is open is therefore
m v ' = m g - k v
or
v ' = g - k/m * v
dv / (g - k/m * v) = dt
- m/k ln | g - k/m * v | = t + c
ln | g - k/m * v | = -k/m * t + c
g - k/m * v = e^(-k/m * t + c) = A e^(-k/m * t), A > 0
v = m g / k - A e^(-k/m * t).
With m = 80 kg and k = 176 N s / m, so that m g / k = 4.5 approx. and k/m = 2.2 approx.. the equation, assuming SI units throughout, is
v = 4.5 - A e^(-2.2 t).
Let t = 0 when the parachute is opened. Then
98 = 4.5 - A e^(-2.2 * 0)
so
A = -93.5.
Thus
v(t) = 4.5 + 93.5 e^(-2.2 * t).
For reference, we note that the velocities at t = 0, 1, 2, 3 and 4 seconds are, respectively, 98, 14.86009530, 5.647931280, 4.627194411, 4.514093542 (all in m/s). So the parachute slows abruptly, with an average velocity during the first second somewhat less than the mean of 98 m/s and 15 m/s implying a distance of less than about 55 m during the first second..
The position function is an integral of the velocity function, so
s(t) = integral ( 4.5 + 93.5 e^(-2.2 * t) dt) = 4.5 t^2 - 42 e^(-2.2 t) + c.
The change in the position function between t = 0 and t = 4 is therefore about 60 meters.
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