Question

Two cars, one driven be Sappho and the other driven be Homer, start side by side at the beginning of a race.?Velocities of the 2 cars during the race are given in the tables below where velocities, v are in miles per hour and times t are in seconds.

A. Use ten trapezoids to approximate the area between the curves y = vs(t) and y = vH(t) (s and H for sappho and homer) on the interval from t = 0 to t = 10. What definite integral are you approximating?

B. If you haven’t done so, convert your answer from part A into units that make good common sense for the physical situation described. Then explain what practical meaning this answer has for the physical situation.

C. Estimate the average velocity of Sappho’s car over the 10-second interval using 5 subintervals and midpoints.

Answer 1

trapeoidal rule
know that Δx=(b-a)/n where the endpoints are x=a and x=b and n=number of subintervals
for this one, b=10 and a=0 and n=10 so (10-0)/10=10/10=1
where f(x) is the function and we start at x=a and end at x=b and starting with the term of x₁ and ending with
`x_n`, the trapezoidal sum is

`\sum\limits^b_a f(x) dx=(\Delta x)/(2)[f(x_1)+2f(x_2)...+2f(x_(n-1))+f(x_n)]`

hmm, just minus the sums and if it is negative, multiply by -1
the integral we aproximating is
`\int\limits^(10)_0 {vS(t)-vH(t)} \, dt`

that is

`(1)/(2)`[0+2(20)+2(35)+2(48)+2(62)+2(75)+2(85)+2(93)+2(99)+2(106)+111]-
`(1)/(2)`[0+2(18)+2(31)+2(43)+2(58)+2(68)+2(79)+2(86)+2(93)+2(95)+96]=59.5



B. since it was positive, sappho's car traveled 59mi farther than homer's car

C. average velocity is all the velocities added together
wait, midpoints?
hmm, ok, so we want 5 intervals
so we take average of each section then average them
5 sections
ah, got it
for midpoint, we don't generate our own data
example, from t=0 to t=2, the midpoint value is 20 because it is at t=1
we do the intervals
0 to 2, 2 to 4, 4 to 6, 6 to 8, 8 to 10
midpoints are 20, 48, 75, 93, 106
so we do
(1/5)(20+48+75+93+106)=68.4
the average velocity is 68.4 mph