1. 5.1 Accumulated Changes Example 1: An objects travels with a velocity of 15 mph. What is the distance traveled after 4 hours t v 15 1234 Distance = area = (4)(15) =60 The velocity is constant and the distance is: S = v.t = (4 ) (15) = 60 miles

2. Example 2: An objects travels for 4 hours with the following velocities: First hour : 15 mph, Second hour: 20 mph, Last 2 hours: 40 mph What is the distance traveled t v 15 1234 Distance = area = 15 + 20 + 80 = 115 miles Total = 115 miles First hour : (15 mph)(1 hour) = 15 miles Second hour: (20 mph)(1 hour) = 20 miles Last 2 hours: (40 mph)(2 hours) = 80 miles A=15 20 40 A=20 A=80

3. Example 3: An objects travels with the following velocity: v = 15t + 10 Find the distance traveled after 4 hours t v 10 1234 25 40 t v 02341 1040557025 55 70 25 40 55 70 Total From Left: (10)(1) + (25)(1) + (40)(1) + (55)(1) =130 Try t = 0,1,2,3,4 hours  = 1 hour

4. Example 3 (Cont.): Using the same data and graph: t v 10 1234 25 40 55 70 25 40 55 70 Total From Left from previous slide: (10)(1) + (25)(1) + (40)(1) + (55)(1) =130 Total From Right: (70)(1) + (55)(1) + (40)(1) + (25)(1) =190 Average: (130+190)/2 =160 miles  = 1 hour

5. t v 10 1234 25 40 55 70 25 40 55 70  = 1 hour n = 4 intervals Note:  is the interval size n is the number of intervals t 1234 25 40 55 70  = 0.5 hour n = 8 intervals 1.5

6. Example 4: An objects travels with the following velocity: v = 2t 2 + 5 Find the distance traveled between t = 0 and t = 6 using n = 4 intervals t v 5 1.534.56 t v 03 61.5 5 9.5 23 45.5 77 Total From Left: (5)(1.5) + (9.5)(1.5) + (23)(1.5) +  = 1.5 hour  = 6 hours / 4 intervals =1.5 hour 9.5 23 45.5 77 =124.5 (45.5)(1.5) +

7. Example 4 (Cont.): Using the same data and graph: t v 5 1.534.56 9.5 23 45.5 77 Total From Left from previous slide: (77)(1.5) + (45.5)(1.5) + (23)(1.5) +  = 1.5 hour =124.5 (9.5)(1.5) + =232.5 Average: (124.5+232.5)/2 =178.5 miles Total From Right: