Explanation:
a) The distance is the integral of the velocity vs. time graph. We can approximate the distance using a left hand Riemann sum. That means for each interval, use the velocity at the beginning of the interval. Don't forget to convert mi/hr to mi/s.
d₁ = (10 s − 0 s) (183.9 mi/hr × 1 hr / 3600 s) = 0.5108 mi
d₂ = (20 s − 10 s) (169.0 mi/hr × 1 hr / 3600 s) = 0.4694 mi
d₃ = (30 s − 20 s) (105.6 mi/hr × 1 hr / 3600 s) = 0.2933 mi
d₄ = (40 s − 30 s) (99.8 mi/hr × 1 hr / 3600 s) = 0.2772 mi
d₅ = (50 s − 40 s) (124.5 mi/hr × 1 hr / 3600 s) = 0.3458 mi
d₆ = (60 s − 50 s) (177.1 mi/hr × 1 hr / 3600 s) = 0.4936 mi
d = 0.5108 + 0.4694 + 0.2933 + 0.2772 + 0.3458 + 0.4936
d = 2.390 miles
b) Do the same as part a, but this time, use a right hand Riemann sum. Instead of using the velocity at the beginning of the interval, use the velocity at the end of the interval.
d₁ = (10 s − 0 s) (169.0 mi/hr × 1 hr / 3600 s) = 0.4694 mi
d₂ = (20 s − 10 s) (105.6 mi/hr × 1 hr / 3600 s) = 0.2933 mi
d₃ = (30 s − 20 s) (99.8 mi/hr × 1 hr / 3600 s) = 0.2772 mi
d₄ = (40 s − 30 s) (124.5 mi/hr × 1 hr / 3600 s) = 0.3458 mi
d₅ = (50 s − 40 s) (177.1 mi/hr × 1 hr / 3600 s) = 0.4936 mi
d₆ = (60 s − 50 s) (175.6 mi/hr × 1 hr / 3600 s) = 0.4878 mi
d = 0.4694 + 0.2933 + 0.2772 + 0.3458 + 0.4936 + 0.4878
d = 2.367 miles
c) The velocity decreases from 0 s to 30 s, but then increases from 30 s to 50 s, and then decreases from 50 s to 60 s.
Since the velocity doesn't consistently increase or decrease, these Riemann sums are neither lower nor upper sums.