Question #5 Question #5 20052005

Truth… or fiction.Truth… or fiction.

Victory over the beast.Victory over the beast.

The ProblemThe ProblemFrustrationFrustration

AnguishAnguish

The Basics (In Paragraph Form)The Basics (In Paragraph Form)

A car is traveling on a straight road. A car is traveling on a straight road. For For 0≤t≤240≤t≤24 seconds, the car’s seconds, the car’s velocity v(t), in meters per second, is velocity v(t), in meters per second, is modeled by the piecewise-linear modeled by the piecewise-linear function defined by the graph on the function defined by the graph on the next slide. For next slide. For t=(0,4)t=(0,4),, v=5t v=5t; for ; for t=(4,16)t=(4,16),, v=20 v=20; and for ; and for t=(16,24)t=(16,24),, v=-(5/2)t+60v=-(5/2)t+60..

Graphical Nature of the BeastGraphical Nature of the Beast

Car in Hyperdrive

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Part A: Where It All BeganPart A: Where It All Began

Find the integral of Find the integral of v(t)v(t) from from (0,24).(0,24).

Using correct units, explain the Using correct units, explain the meaning of the integral.meaning of the integral.

Answering Part A Using LogicAnswering Part A Using LogicFirst off, the integral of First off, the integral of v(t)v(t) from from 0≤t ≤240≤t ≤24 measures the distance traveled, measures the distance traveled, represented by the area under the represented by the area under the velocity graph. Thus, the integral would velocity graph. Thus, the integral would be measured in meters.be measured in meters.

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Car in Hyperdrive

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Answering Part A Using LogicAnswering Part A Using Logic

The exact value of the integral is The exact value of the integral is 360 360 metersmeters as found by breaking the as found by breaking the graph into 3 separate areas as graph into 3 separate areas as defined by their own functions.defined by their own functions.

Part B: Intermediate StagePart B: Intermediate Stage

For each of For each of v’(4)v’(4) and and v’(20),v’(20), find the find the value or explain why it doesn’t exist.value or explain why it doesn’t exist.

Indicate units of measure.Indicate units of measure.

Answering Part B Using Our MindsAnswering Part B Using Our Minds

Because point Because point v’(4)v’(4) is a point where is a point where two separate functions meet with two separate functions meet with two separate slopes aproaching that two separate slopes aproaching that value, there is no limit or for this value, there is no limit or for this instance, no derivative of instance, no derivative of v’(4).v’(4).

Answering Part B Using Our MindsAnswering Part B Using Our Minds

For For v’(20)v’(20) however, this point is only however, this point is only in the domain of one function, that in the domain of one function, that being being v=-(5/2)t+60. v=-(5/2)t+60. In this case, the In this case, the acceleration of the function, also acceleration of the function, also known as the derivative of v, at the known as the derivative of v, at the point of point of v(20),v(20), is is -5/2 meters per -5/2 meters per secondsecond..

Part C: The ClimaxPart C: The Climax

Let Let a(t)a(t) be the car’s acceleration at be the car’s acceleration at time time tt, in meters per second per , in meters per second per second.second.

For For 0<t<240<t<24, write a piecewise-, write a piecewise-defined function for defined function for a(t).a(t).

Answering Part C The Hard WayAnswering Part C The Hard Way

Since the piece-wise function for the Since the piece-wise function for the velocity of the car is defined by as velocity of the car is defined by as t=(0,4)t=(0,4),, v=5t v=5t; for ; for t=(4,16)t=(4,16),, v=20 v=20; ; and for and for t=(16,24)t=(16,24),, v=-(5/2)t+60 v=-(5/2)t+60, the , the acceleration can be found in as a acceleration can be found in as a piece-wise by taking the derivatives piece-wise by taking the derivatives of each of these functions.of each of these functions.

Answering Part C The Hard WayAnswering Part C The Hard Way

Therefore, the acceleration is found Therefore, the acceleration is found as for as for t=(0,4)t=(0,4),, a=5 a=5; for ; for t=(4,16)t=(4,16),, a=0a=0; and for ; and for t=(16,24)t=(16,24),, a=-(5/2). a=-(5/2).

Part D: The FinalePart D: The Finale

Find the average rate of change of v Find the average rate of change of v over the interval 8≤t≤20.over the interval 8≤t≤20.

Does the Mean Value Theorem Does the Mean Value Theorem guarantee a value.guarantee a value.

Answering Part D With FlareAnswering Part D With FlareSince the average rate of change of v is Since the average rate of change of v is defined as the average acceleration over defined as the average acceleration over the interval, you can find the average by the interval, you can find the average by using the slope of the velocity curve.using the slope of the velocity curve.

By multiplying the slope by the time By multiplying the slope by the time interval during each separate region, then interval during each separate region, then adding them together and dividing by the adding them together and dividing by the total time elapsed, you can find the total time elapsed, you can find the average acceleration.average acceleration.

Answering Part D With FlareAnswering Part D With Flare

The equation form is defined as- The equation form is defined as- aa11(t) (from 8≤t≤16)*8 seconds + (t) (from 8≤t≤16)*8 seconds + aa22(t) (from 16≤t≤20)*4 seconds (t) (from 16≤t≤20)*4 seconds /total time (12 seconds)./total time (12 seconds).

Using substitution it is seen as Using substitution it is seen as ((0 ((0 m/sm/s22)*(8 seconds)+(-5/2 m/s)*(8 seconds)+(-5/2 m/s2 2 )*(4 )*(4 seconds))/(12 seconds).seconds))/(12 seconds).

Answering Part D With FlareAnswering Part D With FlareThen the average acceleration over Then the average acceleration over the interval of the interval of 8≤t≤20 8≤t≤20 is is -5/6 m/s-5/6 m/s22..The Mean Value Theorem is defined The Mean Value Theorem is defined as as f’(c)=(f(b)-f(a))/(b-a).f’(c)=(f(b)-f(a))/(b-a).Therefore, it is saying that in this Therefore, it is saying that in this situation, the average acceleration is situation, the average acceleration is defined as defined as f’(c)=(10 m/s–20 m/s)/(20 f’(c)=(10 m/s–20 m/s)/(20 seconds-8 seconds)= -5/6 m/sseconds-8 seconds)= -5/6 m/s22. .

Answering Part D With FlareAnswering Part D With Flare

However, since at the point However, since at the point (16,20)(16,20) there is an instantaneous change in there is an instantaneous change in acceleration, this means it is not acceleration, this means it is not differentiable on the entire interval differentiable on the entire interval making the Mean Value Theorem not making the Mean Value Theorem not applicable, though the correct value applicable, though the correct value can be found using it.can be found using it.

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