Query 09

#$&*

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:

v(0) = 220mph = 968/3 ft/sec

v(4) = 50mph = 220/3 ft/sec

mv’ = -kv

v’ = -kv/m

ʃdv/v = ʃ-kdt/m

ln(v) + c_1 = -kt/m + c_2

ln(v) = -kt/m + C

v = Ae^(-kt/m) ft/sec

968/3 = Ae^(0)

A = 968/3 ft/sec

v = [968e^(-kt/m)]/3 ft/sec

220/3 = [968e^(-4k/3000)]/3

k = 750ln(22/5) lb/sec ≈ 1111.203 lb/sec

v(t) = [968e^(-(750ln(22/5))t/m)]/3 ft/sec

s(t) = ʃv(t)dt = [3872(5/22)^(t/4)]/(3ln(5/22)) ft

s(4) = 197.98 ft

@&

I agree with everything up to about the last step.

750 ln(22/5) / 3000 is close to .3, so I believe our exponential functions are compatible.

I'm not sure I see the exponential function in your integral.

Compare with my integral and see what you think:

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.

As an estimate, an exponential function that decreases from around 300 ft / sec to around 70 ft / sec should average over 100 ft/sec, which for a 4 second interval would imply a distance over 400 ft, disagreeing with both of our solutions.

In any case I can't find any fault with your solution, at least up to the integration.

*@

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): I think I got this problem, including units. I didn’t carry my units through on the type-up in the interest of conciseness and neatness, but on paper I carried my units through my calculations, and I’m nearly 100% certain that I carried them all through correctly.

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Self-critique rating: 3

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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’ = -mg-kv

v’ = -g-kv/m

v’ + kv/m = -g

µ = e^(ʃkdt/m) = e^(kt/m)

µv’ + µkv/m = -gµ

ʃd/dt(µv)dt = ʃ-gµdt

µv + c_1 = -mge^(kt/m)/k + c_2

µv = -mge^(kt/m)/k + C

v = -mg/k + Ae^(-kt/m)

A = v_0 + mg/k

v(t) = -mg/k + (v_0 + mg/k)e^(-kt/m)

Impose v(t) = 0, solve for t:

t = -m/k*ln((mg/k)/(v_0 + mg/k)) s

This is when the projectile will reach its maximum height.

Using the values given:

2.5 = -0.12/k*ln((0.12*9.81/k)/(80+0.12*9.81/k))

k = 0.0975 g/s

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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:

The first step is to find the initial velocity at the time the chute was opened. The terminal velocity can be found using -mg/k.

We find that k, with a 90kg man reaching a terminal velocity of 5m/s, is 176.58 kg/s.

v(t) = mg/k - (v_0 + mg/k)e^(-kt/m)

If we assume that there is no drag on the skydiver in free-fall, then our initial velocity, v_0 = 9.81*10 = 98.1 m/s.

98.1 = 4.55 - Ae^(-2.15t)

A = -93.4

v(t) = 4.55+93.4e^(-2.15t)

s(t) = ʃv(t)dt = 4.55t-43.46e^(-2.15t)

In the span of 4 seconds, the skydiver travels about 61.65m, meaning that the parachute was opened at an altitude of about 61.65m.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK"

&#This looks good. See my notes. Let me know if you have any questions. &#