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Problem Solving Using Polynomial Equations Objective: To solve problems using polynomial equations

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Physics The motion of an object with respect to gravity can be modeled by a quadratic equation When an object is launched upward with an initial velocity v, its height h from the launch point after t seconds is h = vt – 4.9t 2 meters h = vt – 16t 2 feet The motion of an object with respect to gravity can be modeled by a quadratic equation When an object is launched upward with an initial velocity v, its height h from the launch point after t seconds is h = vt – 4.9t 2 meters h = vt – 16t 2 feet

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Example: In celebration a person fires a bullet into the air. The bullet shoots upward at a speed of 350 m/s. A)How long will it be till the person who fired the gun may accidentally shoot himself in the head. B) What is the maximum height the bullet will reach before it comes speeding back to earth This problem was inspired by the hundreds of Iraqis who managed to injure themselves from shooting guns in the air during celebration of various events In celebration a person fires a bullet into the air. The bullet shoots upward at a speed of 350 m/s. A)How long will it be till the person who fired the gun may accidentally shoot himself in the head. B) What is the maximum height the bullet will reach before it comes speeding back to earth This problem was inspired by the hundreds of Iraqis who managed to injure themselves from shooting guns in the air during celebration of various events

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Solution: A)The height at which the bullet will strike the ground is 0 The velocity is 350 m/s h = vt – 4.9t 2 0 = 350t – 4.9t 2 0 = t(350 – 4.9t) t = 0or350 – 4.9t = 0 t = 0ort 71.429 s A)The height at which the bullet will strike the ground is 0 The velocity is 350 m/s h = vt – 4.9t 2 0 = 350t – 4.9t 2 0 = t(350 – 4.9t) t = 0or350 – 4.9t = 0 t = 0ort 71.429 s

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B) What is the maximum height the bullet will reach before it comes speeding back to earth The bullet should reach its maximum height at the halfway point t = 71.429/2 t = 35.7145 h = vt – 4.9t 2 h = 350(35.7145) – 4.9(35.7145) 2 h = 6250 m Note: these types of problems do not take into consideration, friction, or wind resistance and so can sometimes be unrealistic. The bullet should reach its maximum height at the halfway point t = 71.429/2 t = 35.7145 h = vt – 4.9t 2 h = 350(35.7145) – 4.9(35.7145) 2 h = 6250 m Note: these types of problems do not take into consideration, friction, or wind resistance and so can sometimes be unrealistic.

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Area A tennis court is twice as long as it is wide. There is also an extra 12 feet of space outside of the court lines. If the tarp covering the field is 6426 square feet what are the dimensions of the court?

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A = lw A = 6426 ft l = 2w A = lw 6426 = (2w)(w) Add on the border 6426 = (2w + 12 + 12)(w + 12 + 12) 6426 = (2w + 24)(w + 24) A = 6426 ft l = 2w A = lw 6426 = (2w)(w) Add on the border 6426 = (2w + 12 + 12)(w + 12 + 12) 6426 = (2w + 24)(w + 24) 12ft 12ft

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6426 = (2w + 24)(w + 24) 6426 = w(2w + 24) + 24(2w + 24) 6426 = 2w 2 + 24w + 48w + 576 6426 = 2w 2 + 72w + 576 6426 – 6426 = 2w 2 + 72w + 576 – 6426 2w 2 + 72w – 5850 = 0 2(w 2 + 36w – 2925) = 0 2925 = 3 2 5 2 13 6426 = w(2w + 24) + 24(2w + 24) 6426 = 2w 2 + 24w + 48w + 576 6426 = 2w 2 + 72w + 576 6426 – 6426 = 2w 2 + 72w + 576 – 6426 2w 2 + 72w – 5850 = 0 2(w 2 + 36w – 2925) = 0 2925 = 3 2 5 2 13

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Almost Done 2(w 2 + 36w – 2925) = 0 75 -39 = -292575 + -39 = 36 w 2 +75w – 39w – 2925 = 0 w(w + 75) – 39(w + 75) = 0 (w – 39)(w + 75) = 0 w – 39 = 0or w + 75 = 0 w = 39orw = -75 2(w 2 + 36w – 2925) = 0 75 -39 = -292575 + -39 = 36 w 2 +75w – 39w – 2925 = 0 w(w + 75) – 39(w + 75) = 0 (w – 39)(w + 75) = 0 w – 39 = 0or w + 75 = 0 w = 39orw = -75

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Solution w = -75 is discarded because it doesnt make sense in real life. w = 39 l = 2wl = 2 39 = 78 The dimensions of the court are 39 ft. by 78 ft. w = -75 is discarded because it doesnt make sense in real life. w = 39 l = 2wl = 2 39 = 78 The dimensions of the court are 39 ft. by 78 ft.

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