Problem solving using polynomial equations
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Problem Solving Using Polynomial Equations. Objective: To solve problems using polynomial equations. Physics. The motion of an object with respect to gravity can be modeled by a quadratic equation

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Problem Solving Using Polynomial Equations

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Problem solving using polynomial equations

Problem Solving Using Polynomial Equations

Objective: To solve problems using polynomial equations


Physics

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 – 4.9t2 metersh = vt – 16t2 feet


Example

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 earthThis problem was inspired by the hundreds of Iraqi’s who managed to injure themselves from shooting guns in the air during celebration of various events


Solution

Solution:

  • A)The height at which the bullet will strike the ground is 0

  • The velocity is 350 m/s

  • h = vt – 4.9t2

  • 0 = 350t – 4.9t2

  • 0 = t(350 – 4.9t)

  • t = 0or350 – 4.9t = 0

  • t = 0ort ≈ 71.429 s


B what is the maximum height the bullet will reach before it comes speeding back to earth

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 – 4.9t2

  • 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.


Problem solving using polynomial equations

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?


Problem solving using polynomial equations

↕12ft

12ft

12ft

↕12ft

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)


6426 2w 24 w 24

6426 = (2w + 24)(w + 24)

  • 6426 = w(2w + 24) + 24(2w + 24)

  • 6426 = 2w2 + 24w + 48w + 576

  • 6426 = 2w2 + 72w + 576

  • 6426 – 6426 = 2w2 + 72w + 576 – 6426

  • 2w2 + 72w – 5850 = 0

  • 2(w2 + 36w – 2925) = 0

  • 2925 = 32• 52 • 13


Almost done

Almost Done

  • 2(w2 + 36w – 2925) = 0

  • 75 • -39 = -292575 + -39 = 36

  • w2 +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


Solution1

Solution

  • w = -75 is discarded because it doesn’t 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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