Asst 3

course Mth 174

́E~c|N} f\\яyassignment #003

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Physics II

06-17-2007

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06:06:24

query 6.5 #8 Galileo: time for unif accel object to traverse dist is same as if vel was ave of init and final; put into symbols and show why true

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RESPONSE -->

a(t) = (v1+ v2)/2

I'm not sure how to show this is true. This problem really confused me.

** Using s for the distance fallen we can translate Galileo's statement as follows:

t = s / [ (vf + v0)/2 ].

A ball dropped from rest will have velocity function

v = a * t

and position function

s = .5 a t^2.

Both of these results are easily obtained, for the given initial condition that initial velocity is zero, by a straightforward integration.

For given distance of fall s the position function gives us

time of fall t = sqrt(2 s / a).

At this time the velocity function tells us that

Final velocity = a * time of fall = a sqrt(2 s / a) = sqrt( 2 a s).

From this we calculate the average of initial and final velocities and show that at this velocity the time required to fall distance s is the same as above:

The average of initial and final velocities is

Ave of initial and final velocities = (0 + sqrt( 2 a s) ) / 2 = sqrt( a s / 2 ).

If an object travels distance s at this velocity sqrt( a / (2 s) ) then the time required is

Time required at ave of initial and final velocities

= s / sqrt( a s / 2 ) = sqrt(2 s / a),

Which is in agreement with the result obtained from the position function.

Let t be the time of fall. Then initial velocity is 0 and final velocity is a * t, so the average of initial and final velocities is

Ave of init and final vel = (0 + a * t) / 2 = a t.

Then using Galilieo's assumption we show that they are consistent with the result we get from s = 16 t^2.

If an object is dropped from rest and falls for time t it will reach velocity vf = a t = 32 t. So the average of its initial and final velocities will be (vf + v0) / 2 = (32 t + 0) / 2 = 16 t.

The distance fallen is s = 16 t^2.

The time to fall distance s = 16 t^2 at average velocity 16 t is s / t = 16 t^2 / (16 t) = t, which agrees with the time the object was allowed to fall.

A numerical example for a given s:

When s = 100, then, t is the positive solution to 100 = 16 t^2. This solution is t = 2.5.

The velocity function is v = a t. We have to find v when s = 100. When s = 100, we have t = 2.5, as just seen. So v = a t = 32 * 2.5 = 80, representing 80 ft/sec.

Now we know that s = 100, vf = 80 and v0=0. Substituting into t = s / [ (vf + v0)/2 ] we get t = 100 / [ (80 + 0) / 2 ] = 100 / 40 = 2.5.

This agrees with the t we got using s = .5 a t^2. **

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06:06:45

how can you symbolically represent the give statement?

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RESPONSE -->

Again I was confused with this question.

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06:06:49

How can we show that the statement is true?

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RESPONSE -->

Again I was confused with this question.

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06:06:52

How can we use a graph to show that the statement is true?

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RESPONSE -->

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10:22:32

what did you get for the integral and how did you reason out your result?

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RESPONSE -->

u = cos(3t)

du = -3sin(3t)dt

-(1/3)du = sin(3t)dt

int['sqrt(cos(3t) ) * sin(3t)]

-(1/3)int['sqrt(u)]du

-(1/3)*(2/3) u^(3/2) + c

-(2/9) (cos(3t))^(3/2) + c

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10:25:30

what is the antiderivative?

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RESPONSE -->

int[(x^2)(e^(x^(3)+1))dx

u = x^3 + 1

du = 3x^2 dx

(1/3)du = x^2 dx

(1/3) int [e^u] du

(1/3) e^u + c

(1/3) e^(x^3 + 1) + c

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10:25:43

What substitution would you use to find this antiderivative?

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RESPONSE -->

u = x^3 + 1

du = 3x^2 dx

(1/3)du = x^2 dx

(1/3) int [e^u] du

(1/3) e^u + c

(1/3) e^(x^3 + 1) + c

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10:38:35

what is the antiderivative?

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RESPONSE -->

I'm not sure what to use as u. I tried t+1 and t^2.

when I used

u= t^2

du = 2tdt

(1/2)tdu = dt

(1/2)t int[((t+1)^2)/u) du

1/2 t + ln (t^2)

I got stuck here. This isn't right, but I can't get the correct answer.

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10:38:46

What substitution would you use to find this antiderivative?

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RESPONSE -->

That's where i'm confused.

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10:42:30

What did you get for the definite integral?

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RESPONSE -->

u = t + 7

du = dt

int[u^-2]du

1/u

1/(t+7) (1, 3)

1/10 - 1/8 = -1/40

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10:42:39

What antiderivative did you use?

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RESPONSE -->

int[u^-2]du

1/u

1/(t+7) (1, 3)

1/10 - 1/8 = -1/40

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10:42:46

What is the value of your antiderivative at t = 1 and at t = 3?

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RESPONSE -->

1/(t+7) (1, 3)

1/10 - 1/8 = -1/40

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10:44:25

What were the populations in 1990 and 2000?

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RESPONSE -->

1990 population is 5.3billion

2000 population is 6.1 billion

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10:47:36

What is the average population between during the 1990's and how did you find it?

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RESPONSE -->

int 5.3e^(0.014t) (0, 10)

int 5.3e^(0.014*0) = 5.3 billion

int 5.3e^(0.014*10) = 6.1 billion

6.1 billion - 5.3 billion = .8 billion

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10:49:21

What is the value of your antiderivative at t = 1 and at t = 3?

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RESPONSE -->

0.8 billion work on last problem.

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"

See my note on the first problem.

You do not appear to be using the program correctly. Please run it again. You can easily copy and paste the responses you've inserted here.

From what I can see your work looks pretty good. However I need to see everything so I can see the entire flow of questions and answers. Many of your responses did not contain the questions, and I cannot efficiently look back and forth among multiple documents to get the information I need to assess your work.

&#To run the program correctly you click on the Next Question/Solution button (top left), enter your response in the Answer box (the box at top right), click on the Enter Response button (above the top right box), then again on the Next Question/Solution botton (top left) to see the solution, and finally enter your response or self-critique in the Answer box and click on Enter Response.

You then start the process over with the next question, clicking on the Next Question/Solution button at top left, etc.

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