#$&*
course MTH-279
06/25 around 11:40AM.
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?YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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
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You might be better off to express 220 mph in feet/second, so that your distance will come out in feet.
However as long as you are careful about units this will work.
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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 = -34.5
Which makes the equation:
v(t) = 220e^(-0.37t)
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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 feet
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It's not clear that the units of this calculation come out in feet.
220 mph is about 323 ft / sec. Using this value we would get 670 feet.
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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:
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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The antiderivative of 1 / u is ln | u |, so you have
| kv/m + g | = Ce^(-kt/m)
The solution
kv/m + g = -Ce^(-kt/m)
is also possible, leading to
v = m g / k - C e^(-k t / m).
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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
???? Lost after this step???
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I believe the equation for t_max, the time at which v = 0, will be
t_max = -1/k ln | 1 - k v_0 / g |
This equation cannot be solved exactly.
Newton's Method indicates a maximum near k = -.2.
This yields a velocity function which decreases at a decreasing rate (i.e., which is concave up) passing through (0, 80 m/s) and (2.5 s, 0).
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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
mv' = F_net = kv + mg
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(t) = Ce^(-kt/m) - mg/k
Note the negative sign in the exponent.
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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
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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 - C e^(-2.2 t).
Let t = 0 when the parachute is opened. Then
98 = 4.5 - C e^(-2.2 * 0)
so
C = -93.5.
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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 m
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I get a final result of about 60 metes.
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&#Good responses. See my notes and let me know if you have questions. &#