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course MTH 279
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:
We know that dv/dt = F, where F is the sum of all forces.
We also know that by substituting in F = -kv, we can get: dv/dt = -mg -kv. since we know mv' + kv = -mg, we can rearrange the equation to be v' + (k/m) v = -g. Then using an integrating factor e^(k/m)t, we can multiply
to by that and then integrate to get v(t) = -(mg/k) + Ce^(-k/m)t. Since we have the initial conditions to be t = 0, v(0) = 220, we can plug in the initial conditions to find C: C = v(0) + (mg/k).
Therefore we simplify our equation to be v(t) = -(mg/k)(1-e^(-k/m)t) + v(0)e^(-k/m)t. Now we can use our other conditions g = 32, m = 3000, t = 4, v(0) = 220, and v(t) = 50, to find k.
Since the car is driving horizontally, we can eliminate the first part of the equation because there is no pull from gravity. Therefore our equation is v(t) = v(0)e^(-k/m)t. Plugging in our conditions we get k = 1111. The car will travel about .4 miles.
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Your solution is good but I'm not sure if you accounted fully for the units of the given information (e.g., velocities are in mph but time is in seconds), so your final result for the distance is clearly off. Even at 220 mph the car wouldn't travel .4 miles in 4 seconds.
Check the solution below:
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 4 s)
= -1/.31 s * 323 ft / s * (e^(-.31 s^-1 * 4 s) - e^0)
= 1000 ft * (1 - .28) = 720 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?
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Your solution:
The highest point occurs when v(t) = 0. Therefore we would set up our equation v(t) and then solve for when it equals 0 to find t.
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The acceleration is - g. You can and should set up and solve the equation and evaluate the integration constant in terms of v_0, k and g.
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There is, or should be, an additional question:
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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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.
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Your solution:
What are you looking for? Not seeing a question or something to solve?
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The rest of the question reads
"At what altitude was the parachute opened?"
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The last two questions were indeed missing the question to be answered. However it would have been possible to set up and solve the differential equation for each question.
I've modified the Query to include the entire question, and I'll check subsequent queries to see if something similar has happened there.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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