2. Question #5 2005 Truth… or fiction.

3. Victory over the beast.

4. The Problem FrustrationAnguish

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

6. Graphical Nature of the Beast

7. Part A: Where It All Began Find the integral of v(t) from (0,24). Using correct units, explain the meaning of the integral.

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

9. Nick Scherer Rocks!

10. Answering Part A Using Logic The exact value of the integral is 360 meters as found by breaking the graph into 3 separate areas as defined by their own functions.

11. Part B: Intermediate Stage For each of v’(4) and v’(20), find the value or explain why it doesn’t exist. Indicate units of measure.

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

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

14. Part C: The Climax Let a(t) be the car’s acceleration at time t, in meters per second per second. For 0<t<24, write a piecewise- defined function for a(t).

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

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

17. Part D: The Finale Find the average rate of change of v over the interval 8≤t≤20. Does the Mean Value Theorem guarantee a value.

18. Answering Part D With Flare Since the average rate of change of v is defined as the average acceleration over the interval, you can find the average by using the slope of the velocity curve. By multiplying the slope by the time interval during each separate region, then adding them together and dividing by the total time elapsed, you can find the average acceleration.

19. Answering Part D With Flare The equation form is defined as- a 1 (t) (from 8≤t≤16)*8 seconds + a 2 (t) (from 16≤t≤20)*4 seconds /total time (12 seconds). Using substitution it is seen as ((0 m/s 2 )*(8 seconds)+(-5/2 m/s 2 )*(4 seconds))/(12 seconds).

20. Answering Part D With Flare Then the average acceleration over the interval of 8≤t≤20 is -5/6 m/s 2. The Mean Value Theorem is defined as f’(c)=(f(b)-f(a))/(b-a). Therefore, it is saying that in this situation, the average acceleration is defined as f’(c)=(10 m/s–20 m/s)/(20 seconds-8 seconds)= -5/6 m/s 2.

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

22. Calculus Rocks Yay Mrs. Scherer Calculus Rocks