group Homework 12 - We will divide the graph into 8...

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Homework 12 5/3 #34 We know: The cyclist begins at point 5 miles from the lake. Positive velocities (v(t)) take her away from the lake while negative velocities (-v(t)) take her toward the lake. We can see by the graph that her distance traveled at a positive velocity is much greater than her distance traveled at a negative velocity. (Because the area between the curve and the axis is much greater for when the graph of her velocity is positive than when the graph of her velocities is negative.) Therefore, her net displacement will be in the positive direction (away from the lake). She will be the farthest away from the lake at the END of her trip because her net displacement is positive (this occurs at t=1, one hour). Now we need to find how far she is from her starting point: We will divide the graph into 6 subintervals so n=6 and Δt=(1/6). = - = - - - = - = 6 1 1 6 1 1 1 ) )( ( ) )( ( n n n n n n n n t t t v um RightHandS t t t v m LeftHandSu 6/1 #18. We know: g(0)=50 g’ is graphed below: Using this graph we can form a chart for g(x).
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Unformatted text preview: We will divide the graph into 8 divisions with a Δx of 5. (these divisions are marked by dashed lines on the graph). First we can find the area between the curve and the x-axis from x=0 to x=5. The first area is given by a trapezoid so we can divide the area into a rectangle and a triangle. The rectangle portion will be -10*5 which equals -50 and the triangle part will be (1/2)((-15-(-10))*5) which equals -25 so the area of the first trapezoid will be -25. So, at point 5, g(x) will be 50-25=25. We will use this same method for the rest of the divisions. For the division at x=10 we have a larger trapezoid. So the rectangle area is (-10*10)=-100 and the triangle area is (1/2)((-20-(-10))*10)=-50 so the total area is -150. So g(10) is 25-150=-125. …… x 0 5 10 15 20 25 30 35 40 g(x) 50 25 -125 Should I keep going with this??? (10,-20)-10 g’(x) 40 15 (20,10) x...
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