1. Position, Velocity, and Acceleration By Han Kim Period 1

2. Interpreting the Velocity Graph Particle problems involving graphs usually give the graph or a x/y chart of a velocity (F’(x)). t (seconds) = x axis u/sec (or min/hr depending on the time interval) = y axis From the Velocity graph you can derive a number information about the acceleration and the position. The next following sides are examples

3. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps Find v(3) and v(-8)

4. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Find v(3) and v(-8) find y when t=3 & -8 (you can just look at the graph since the graph is v(t) Click for further steps

5. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Find v(3) and v(-8) find y when t=3 & -8 (you can just look at the graph since the graph is v(t) when x=3, y=7 ---- v(3) = 7 when x=-8, y =-6 ---- v(-8) = -6 Click for further steps

6. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Find a(2) & a(-4) Click for further steps

7. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Find a(2) & a(-4) NOTE: Acceleration is the derivative of the velocity-therefore making acceleration the slope of the velocity. By calculating the slope of the line in a velocity graph you can find the acceleration Click for further steps

8. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Find a(2) & a(-4) NOTE: Acceleration is the derivative of the velocity-therefore making acceleration the slope of the velocity. By calculating the slope of the line in a velocity graph you can find the acceleration At 2 the slope = ((Y1)-(Y2))/((X1)-(X2)) line= from (3,7) to (1,0) At -4 the slope = ((Y1)-(Y2))/((X1)-(X2)) line= from (-3,4) to (-6,-2) Click for further steps

9. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Find a(2) & a(-4) NOTE: Acceleration is the derivative of the velocity-therefore making acceleration the slope of the velocity. By calculating the slope of the line in a velocity graph you can find the acceleration At 2 the slope = ((Y1)-(Y2))/((X1)-(X2)) line= from (3,7) to (1,0) (7-0)/(3-1)= 7/2 a(2) = 7/2 or 3.5 At -4 the slope = ((Y1)-(Y2))/((X1)-(X2)) line= from (-3,4) to (-6,-2) (4-(-2))/(-3-(-6))= 6/3 = 2 a(-4) = 2 Click for further steps

10. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps Draw the graph of Acceleration in./min^2 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4

11. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps Draw the graph of Acceleration From -10 min to -8 min the slope is -3 in./min^2 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4

12. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps Draw the graph of Acceleration From -8 min to -7 min the slope is 4 in./min^2 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4

13. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps Draw the graph of Acceleration From -7 min to -6 min the slope is 0 (no acceleration) in./min^2 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4

14. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps Draw the graph of Acceleration From -6 min to -3 min the slope is 2 From -3 min to -1 min the slope is -2 From -1 min to 0 min the slope is -3 From 0 min to 1 min the slope is 3 in./min^2 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4

15. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps Draw the graph of Acceleration From 1 min to 3 min the slope is 3.5 or 7/2 From 3 min to 6 min the slope is -3.5 or -7/2 From 6 min to 7 min the slope is 0 From 7 min to 10 min the slope is 2 in./min^2 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4

16. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps Draw the graph of Acceleration The Graph of the acceleration is a piecewise function where each interval the y value changes The x value is undefined (DNE). From this, you can easily find acceleration at a given (t) This graph was made only to show how it can be done using the velocity graph. By knowing how to do this you will be able to easily find acceleration from v(t) graph. in./min^2 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4

17. t (minutes) in./min^2 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 Graph of a(t) When is acceleration positive? negative? Zero?

18. t (minutes) in./min^2 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 Graph of a(t) When is acceleration positive? negative? Zero? By creating this graph previously we can easily tell where acceleration is positive, negative or zero.

19. t (minutes) in./min^2 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 Graph of a(t) When is acceleration positive? negative? Zero? By creating this graph previously we can easily tell where acceleration is positive, negative or zero. Positive above the x axis Zero on the x axis Negative below the x axis

20. t (minutes) in./min^2 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 Graph of a(t) When is acceleration positive? negative? Zero? By creating this graph previously we can easily tell where acceleration is positive, negative or zero. Positive above the x axis (-8,-7),(-6,-3),(0,1),(1,3),(7,10) Zero on the x axis (-7,-6),(6,7) Negative below the x axis (-10,-8),(-3,-1),(-1,0),(-3,6)

21. t (minutes) in./min^2 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 Graph of a(t) What is the Maximum and Minimum Acceleration?

22. t (minutes) in./min^2 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 Graph of a(t) What is the Maximum and Minimum Acceleration? look at your a(t) graph or calculate the greatest slope in the v(t) graph.

23. t (minutes) in./min^2 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 Graph of a(t) What is the Maximum and Minimum Acceleration? look at your a(t) graph or calculate the greatest slope in the v(t) graph. From this graph we can see that the greatest acceleration/slope of a v(t) graph is 4

24. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps When does the particle move to the left or the right? S(t) number line

25. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps When does the particle move to the left or the right? Velocity graph is the derivative of position- v(t)=s’(t) so any x on the graph that has a positive y value will increase (move to the right) while any x on the graph that has a negative y will decrease (move to the left) S(t) number line

26. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps When does the particle move to the left or the right? Velocity graph is the derivative of position- v(t)=s’(t) so any x on the graph that has a positive y will increase (move to the right) while any x on the graph that has a negative y will decrease (move to the left) Critical points are x intercepts (roots) since this is a s’(t) graph. x=-5, -1, 1, 5, 9 -5159 -+++-- S(t) number line - - - + + +

27. Given: Graph of V(t) and S(0)=2 t (minutes) Click for further steps When does the particle move to the left or the right? Velocity graph is the derivative of position- v(t)=s’(t) so any x on the graph that has a positive y will increase (move to the right) while any x on the graph that has a negative y will decrease (move to the left) Critical points are x intercepts (roots) since this is a s’(t) graph. x=-5, -1, 1, 5, 9 Increasing where Y is positive = (-5,-1), (1,5), (9, infinity) Decreasing where Y is negative = (infinity,-5), (-1, 1), (5,9) -5159 -+++-- S(t) number line 10 in./min 2 4 6 8 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 - - - + + +

28. Given: Graph of V(t) and S(0)=2 t (minutes) Click for further steps Where is the particle changing directions? (hint: use the number line) -5159 -+++-- S(t) number line 10 in./min 2 4 6 8 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 - - - + + +

29. Given: Graph of V(t) and S(0)=2 t (minutes) Click for further steps Where is the particle changing directions? (hint: use the number line) A particle changes direction at every (t) where the position shifts from increasing to decreasing or decreasing to increasing. Every (t) where it shifts from – to +. -5159 -+++-- S(t) number line 10 in./min 2 4 6 8 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 - - - + + +

30. Given: Graph of V(t) and S(0)=2 t (minutes) Click for further steps Where is the particle changing directions? (hint: use the number line) A particle changes direction at every (t) where the position shifts from increasing to decreasing or decreasing to increasing. Every (t) where it shifts from – to +. Particle changes direction at t=-5, -1, 1, 5, 9 (5 times) -5159 -+++-- S(t) number line in./min 2 4 6 8 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 - - - + + + 10

31. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps When does the particle speed up? Slow Down?

32. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps When does the particle speed up? Slow Down? A particle is speeding up or slowing down if both velocity and acceleration are going the same direction (+ or -).

33. Given: Graph of V(t) and S(0)=2 t (minutes) in./min Click for further steps When does the particle speed up? Slow Down? A particle is speeding up or slowing down if both velocity and acceleration are going the same direction (+ or -). The red + and – show where the velocity is increasing or decreasing. 2 4 6 8 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 - - - + + 10

34. Given: Graph of V(t) and S(0)=2 t (minutes) in./min Click for further steps When does the particle speed up? Slow Down? A particle is speeding up or slowing down if both velocity and acceleration are going the same direction (+ or -). The red + and – show where the velocity is increasing or decreasing. Acceleration is increasing where slope is positive – blue + and – show where the slope is positive or negative. 2 4 6 8 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 - - - + + 10 - - - + + +

35. Given: Graph of V(t) and S(0)=2 t (minutes) in./min Click for further steps When does the particle speed up? Slow Down? A particle is speeding up or slowing down if both velocity and acceleration are going the same direction (+ or -). The red + and – show where the velocity is increasing or decreasing. Acceleration is increasing where slope is positive – blue + and – show where the slope is positive or negative. The green box is where speed increases. Although velocity is sometimes decreasing and has a negative slope, because both are negatives the speed increases. Orange box shows where the speed decreases – the interval where either V is positive and A is negative or visa versa. 2 4 6 8 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 - - - + + 10 - - - + + + +

36. Given: Graph of V(t) and S(0)=2 t (minutes) in./min Click for further steps When does the particle speed up? Slow Down? A particle is speeding up or slowing down if both velocity and acceleration are going the same direction (+ or -). The red + and – show where the velocity is increasing or decreasing. Acceleration is increasing where slope is positive – blue + and – show where the slope is positive or negative. The green box is where speed increases. Although velocity is sometimes decreasing and has a negative slope, because both are negatives the speed increases. Orange box shows where the speed decreases – the interval where either V is positive and A is negative or visa versa. 2 4 6 8 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 - - - + + 10 - - - + + + + Speed is increasing: (-10,-8), (-5,-3), (-1,0), (1,3), (5,6), (9,10) Speed is decreasing: (-8,-7), (-6,-5), (-3,-1), (0,1), (3,5), (7,9)

37. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps When is the particle at rest for an instant? More than an instant?

38. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps When is the particle at rest for an instant? More than an instant? A particle is at rest when the velocity is equal to 0 meaning there is no change in speed – it is still (at rest)

39. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps When is the particle at rest for an instant? More than an instant? A particle is at rest when the velocity is equal to 0 meaning there is no change in speed – it is still (at rest) Given the velocity graph- t is resting where v(t)=0 t = -5, -1, 1, 5, 7

40. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps When is the particle at rest for an instant? More than an instant? A particle is at rest when the velocity is equal to 0 meaning there is no change in speed – it is still (at rest) Given the velocity graph- t is resting where v(t)=0 t = -5, -1, 1, 5, 7 The particle is never at rest for more than an instant because v(t) never equals zero simultaneously but only at points.

41. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps What is the total distance traveled by the particles from -10 to 10 minutes.

42. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps What is the total distance traveled by the particles from -5 to 5 minutes. total distance is the absolute value of every area under each curve added together. NOTE: its only from -5 to 5 A F E D C B

43. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps What is the total distance traveled by the particles from -5 to 5 minutes. total distance is the absolute value of every area under each curve added together. NOTE: its only from -5 to 5 lBl + lCl + lDl = total distance traveled. In order to find area use simple triangles = 1/2bh B= ((1/2)(2)(4)) + ((1/2)(2)(4)) = 8 u^2 C= ((1/2)(1)(3)) + ((1/2)(1)(3)) = 3 u^2 D= ((1/2)(2)(7)) + ((1/2)(2)(7)) = 14 u^2 A F E D C B 8 + 3 + 14

44. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps What is the total distance traveled by the particles from -5 to 5 minutes. total distance is the absolute value of every area under each curve added together. NOTE: its only from -5 to 5 lBl + lCl + lDl = total distance traveled. In order to find area use simple triangles = 1/2bh B= ((1/2)(2)(4)) + ((1/2)(2)(4)) = 8 u^2 C= ((1/2)(1)(3)) + ((1/2)(1)(3)) = 3 u^2 D= ((1/2)(2)(7)) + ((1/2)(2)(7)) = 14 u^2 Add all the areas together to get the total distance from -5 to 5. Total distance = 25 inches A F E D C B 8 + 3 + 14

45. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps Draw the graph of Position on the interval (-5,5) in. 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4

46. Given: Graph of V(t) and S(0)=2 in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps in. 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 t (minutes) Draw the graph of Position on the interval (-5,5) the Given states that when t=0, s(t)=2 To graph a s(t) graph all you need to do is know the area under the curve and whether or not it is Increasing or decreasing.

47. Given: Graph of V(t) and S(0)=2 in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps in. 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 t (minutes) Draw the graph of Position on the interval (-5,5) First find all the area under the curve We can use what we found previously D C B 8 3 14 - + +

48. Given: Graph of V(t) and S(0)=2 in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps in. 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 t (minutes) Draw the graph of Position on the interval (-5,5) First find all the area under the curve We can use what we found previously From (0,1) the area is 1.5 and because it is under the x-axis it is going left (decreasing) In the s(t) graph. So from (0,1) the graph Decreases by 1.5. NOTE: the length of the line does not matter, it is the y value that shifts completely down 1.5. D C B 8 3 14 - + +

49. Given: Graph of V(t) and S(0)=2 in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps in. 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 t (minutes) Draw the graph of Position on the interval (-5,5) From (1,5) the area is 14 and because it is above the x-axis it is going right (increasing) In the s(t) graph. So from (1,5) the graph Increases by 14. This changes y from.5 to 14.5 D C B 8 3 14 - + +

50. Given: Graph of V(t) and S(0)=2 in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps in. 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 t (minutes) Draw the graph of Position on the interval (-5,5) From (-1,0) it is tricky – you must work Backwards. In the interval it deceases By 1.5 however it decrease from t=-1 down To t=0. now at 3.5 From (-5,-1) it increases by 8. However It increases starting at t=-5 then 8 up to t=-1 now at -4.5 D C B 8 3 14 - + +

51. Given: Graph of V(t) and S(0)=2 in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps in. 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 t (minutes) Draw the graph of Position on the interval (-5,5) This is the graph of s(t) from (-5,5) Now it’s easy to answer the following Find when the particle is farthest to the right On the interval (-5,5) Given s(0)=2 how far to the right (maximum) does it go in the interval (-5,5). D C B 8 3 14 - + +

52. Given: Graph of V(t) and S(0)=2 in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps in. 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 t (minutes) Draw the graph of Position on the interval (-5,5) This is the graph of s(t) from (-5,5) Now it’s easy to answer the following Find when the particle is farthest to the right On the interval (-5,5) When t=5 Given s(0)=2 how far to the right (maximum) does it go in the interval (-5,5). Maximum is y value when t=5 which is 14.5 D C B 8 3 14 - + +

53. Given: Graph of V(t) and S(0)=2 in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps in. 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 t (minutes) Draw the graph of Position on the interval (-5,5) This is the graph of s(t) from (-5,5) Now it’s easy to answer the following At what (if any) times does s(t) = 3.5 D C B 8 3 14 - + +

54. Given: Graph of V(t) and S(0)=2 in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps in. 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 t (minutes) Draw the graph of Position on the interval (-5,5) This is the graph of s(t) from (-5,5) Now it’s easy to answer the following At what (if any) times does s(t) = 3.5 s(t) = 3.5 when t=-1 and 2 D C B 8 3 14 - + +

55. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps Find the Average Acceleration and Average Velocity from (-6,6)

56. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps Find the Average Acceleration and Average Velocity from (-6,6) Average Acceleration just need to know the formula. v(maximum x) – v(minimum x) Maximum x – minimum x Max= the largest t on the interval = 6 Min= the smallest t on the interval = -6

57. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps Find the Average Acceleration and Average Velocity from (-6,6) Average Acceleration just need to know the formula. v(maximum x) – v(minimum x) Maximum x – minimum x Max= the largest t on the interval = 6 Min= the smallest t on the interval = -6 V(6) – V(-6) = -4 – (-2) = -2 = -1 6-(-6) 12 12 6

58. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps Find the Average Acceleration and Average Velocity from (-6,6) Average Velocity Average value theorem: (1/a-b)∫a to b (v(t))dt or simply one over total interval multiplied by the final distance traveled.

59. Given: Graph of V(t) and S(0)=2 t (minutes) in./min 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 -6 Click for further steps Find the Average Acceleration and Average Velocity from (-6,6) Average Velocity Average value theorem: (1/a-b)∫a to b (v(t))dt or simply one over total interval multiplied by the final distance traveled. (1/12) = interval (1/12) [Area under each curve subtracted if under x axis and added if above x axis] (1/12)[-1 + 8 – 3 + 14 – 2] = 16/12 or 4/3

60. Click for further steps Find the Average Position of the particle from interval (-5,5) Given: Graph of S(t) in. 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 t (minutes)

61. Click for further steps Find the Average Position of the particle from interval (-5,5) By drawing the graph of s(t) you can find average position by adding up all the area under the curve (taking in account if it is above or below the x-axis then multiplying it to one over the total interval (1/10) Average Value: (1/a-b)∫a to b (S(t))dt Given: Graph of S(t) in. 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 t (minutes)

62. Given: Graph of S(t) Click for further steps in. 2 4 6 8 10 -2 -4 -6 -8-10 2 4 6 -2 -4 t (minutes) Find the Average Position of the particle from interval (-5,5) By drawing the graph of s(t) you can find average position by adding up all the area under the curve (taking in account if it is above or below the x-axis then multiplying it to one over the total interval (1/10) Average Value: (1/a-b)∫a to b (S(t))dt (1/10)[Area under the curve with – or +] (1/10)[-3+6+28] = 31/10