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course Mth 279
6/30 9:51 pm
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:
I feel like I don't have enough values for variables
v' + k/m v = -g
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The car is not falling, so g is not relevant. Your equation is simply
v ' + k / m v = 0.
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You need to solve this equation before you substitute anything.
The equation is first-order linear and homogeneous.
It is also separable.
It can be solved by either or, preferably (for the extra insight it provides), by both of those methods.
Having obtained your solution you can plug in the numbers for the symbols, and use the conditions you have correctly stated:
v(0) = 220 mph
v(4) = 50 mph
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v' is the change in v over the change in t. Since the car slowed 170 mph (249.333 feet per second) I can substitute
249.3333/4 + k/3000 v = -3.2
However, I have no way to solve for k. The only way to solve for the distance would be to take the integral of the velocity
62.333 + k/3000 v = -3.2
v = -177400/k
integrating
x = -177400/k * t + C
But that just adds another variable I can't solve for
v(0) = 220 mph
v(4) = 50 mph
I think there should be a t somewhere in my equation for v(t)
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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:
How do I know whether the drag force is v or v^2???
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You are told that the drag force has magnitude k v.
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m dv/dt = -mg - kv
v' + k/m v = -g
dv/dt * dt/dx = dv/dx
dv/dx + k/m v = -g
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dv/dt * dt/dx = dv/dx is true, but this doesn't say that dv/dt = dv/dx.
YOu have replaced v ' = dv/dt with dv/dx, with no change to the rest of the equation.
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when the velocity = 0, the projectile will be at maximum height
dv/dx + k/m (0) = -g
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Your initial conditions must be substituted after the equations is solved, not before. The equation works for functions of t, not for the constant values of those functions at some specific value of t.
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dv/dx = -g
dv = -gdx
integrate
v = -gx + C
v = -gx + v_0
I don't really know how to solve for k, time *and* distance at the same time. Or separately for that matter.
Does the fact that the ball is thrown upwards make this problem different from the first one???
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By using -g you have declared the positive direction to be up. so the initial velocity must be positive.
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You need to return to the equation
m dv/dt = -mg - kv
and solve the equation for v(t). The equation is separable and the integration requires just straightforward u substitution.
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Given Solution:
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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:
In the equation mv' +kv = -mg, which term is terminal velocity??? Or how to I solve for that???
mv' is the same as mass * acceleration, which is force. I don't know what the kv is. Then the -mg is the force of gravity pointing downwards.
So kv is the air resistance. And when I solve for an equation of v(t), the terminal velocity is v(t) as it approaches infinity???
Not too sure how knowing that is helpful.
Two part problem?
Part one: falling freely with air resistance slowing him down
Part two: falling with chute, both chute and air resistance slowing him down.
Part one:
v' + k/m v = -g
dt = 10 s
m = 82 kg
v' + k/82 v = -9.8
I still don't know how to solve for k ???
In the book example, it kind of just disappears out of the equation.
Part two:
dt = 4 s
terminal velocity = 5 m/s. Is that before or as he's hitting the ground????
How did his mass change to 90 kg???
v' + k/90 v = -9.8
I still don't know k or how to find height, or time.
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The first thing you have to do is solve the equation
v' + k/82 v = -9.8
Then you can formulate the given conditions and find the constants, which will be k and the integration constant you get when solving.
This equation is first-order linear nonhomogeneous, and it is also separable. If you can, I recommend solving it both ways and reconciling the two solutions.
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Given Solution:
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Your solutions are often a little out of order. On at least a couple of problems you've tried to impose the given conditions before solving the equation.
In most cases you do have a correct equation. Your first step will be to solve the equation, then to impose the conditions in order to evaluate the constants.
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