#$&*
course MTH 279
5:01 pm 3/01
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:
v_0 = 323 ft/s
v_f = 73.3 ft/s
`d v = -249.3 ft/s
`d t = 4 s
`d v/ `d t = a_ave = -249.3/4 = -62.425 ft/s/s
x = 323*4 - 1/2(62.425)(16) = 792.6 ft
m dv/dt = -kv
m/v dv= -dt
m ln v = -t + C
v = e^((C-t)/m)
v(0)= 323 --> C = 17333
v(4) = 73.3 --> k = 1112.3
@& m x '' = -k x ' so
m x '' + k x ' = 0.
v = x ' so
m v ' + k v = 0
and
v ' / v = -k / m
so
ln | v | = -k / m * t + c
and
v = e^(-k / m * t + c)
or
v = A e^(-k /m * t) , A > 0.
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.
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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:
m dv/dt = -mg- kv
m/(-mg-kv) dv = dt
Integrate:
-m/k ln (-mg-kv) = t + C
Solve for v:
ln (-mg-kv)= C -k/m*t
-mg-kv = Ce^(-k/m*t)
v = (mg -Ce^(-k/m*t))/k
v = v_0 @ t= 0
C= mg-kv_0
v = (mg-(mg-kv_0)(e^(-k/m*t))/k
h_max @ v= 0
mg = (mg-kv_0)e^(-k/m*t)
mg/(mg-kv_0) = e^(-k/m*t)
-k/m*t = ln (mg/(mg-kv_0)
t_max= -m/k ln (mg/(mg-kv_0)
m = 0.00012 t = 2.5
Actually, there doesn't seem to be anyway to solve for k explicitly. All we can do is sub in and simplify some.
v = (0.00012*9.8)-(0.00012*9.8 - kv_0)e^(-k(2.5/0.00012))/k =0
0.001176= (0.001176-kv_0)e^-k(20833)
After working the next problem I found the easier, neater solution to this question but still can't solve for k
v = -mg/k + (v_0+mg/k)e^(-k/m*t)
@& Good.
The problem as stated was supposed to also give you the initial velocity, which is 80 m/s.
This gives you an equation with only k as the unknown.
However the equation gives k as an implicit solution, and you can't solve explicity for k.
The equation would be
80 = .00012 * 9.8 / k + (80 + .00012 * 9.8 / k) e^(-k/.00012 *2.5)
from which you can get the equation
.00012 * 9.8 / k + (80 + .00012 * 9.8 / k) e^(-k/.00012 *2.5) - 80 = 0.
This is of the form
f(t) = 0
with
f(t) = 00012 * 9.8 / k + (80 + .00012 * 9.8 / k) e^(-k/.00012 *2.5)- 80 .
Graphing this function vs. k we find that k = 1.5, roughly.
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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:
Find k with lim v = 5 = mg/k --> k =176.4
v_0 =9.8(10)= 98 m/s
dv/dt +176/82 = 9.8
dv/dt+176.4/82 = 9.8
v = 4.56 + Ce^(-2.13t)
C= 93.4
v= 4.56 + 93.4e^(-2.13t)
v = dx/dt --> int v dt = x
x =4.56t+93.4/(-2.13)e^(-2.13t)
x(4) = 18.23 m
@& You need to calculate x(4) - x(0).
x(0) isn't zero, unless you include an integration constant c, which for x(0) = 0 wouldn't be zero.
For your function x(0) would be about 40 meters, and your integral would be about 60 meters.*@
This seems way off since you would never open a chute at 18 m and I think it's my k value that's way too high. The diver has fallen 490 m before opening the chute and I just don't see 18.23 m as being enough to slow down to a safe touchdown speed. Of course on the other hand you would be going v= 1.27 m/s which does seem more reasonable and maybe it's just the 'theoretical' values used here that are off.
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Given Solution:
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Self-critique rating:"
@& Very well done, but with what appears to be one error in detail (causing your skydiver to open his chute way too close to the ground) and some justifiable confusion on the solution of the second problem for k (the equation is not solvable, except by approximation).
Check my notes.*@