Unformatted text preview: Exercises 1.2
where the acceleration constant a has diﬀerent values for Kevin and for Alison. Kevin covers the last
1 4 of the full distance, L, in 3 seconds. This means Kevin’s acceleration, aK , is 2L − aK 2(3L/4) , aK determined by: tK − t3/4 = 3 = where tK is the time it takes for Kevin to ﬁnish the race. Solving this equation for aK gives, 2 √ 2 − 3/2 L. aK = 9 Therefore the time required for Kevin to ﬁnish the race is given by: tK = √ 2L 2−
2 =√ L/9 3 2− 3/2 √ √ 2 = 12 + 6 3 ≈ 22.39 sec. 3/2 Alison covers the last 1/3 of the distance, L, in 4 seconds. This means Alison’s acceleration, aA , is found by: tA − t2/3 = 4 = 2L − aA 2(2L/3) , aA where tA is the time required for Alison to ﬁnish the race. Solving this equation for aA gives 2 √ 2 − 4/3 L. aA = 16 Therefore the time required for Alison to ﬁnish the race is given by: tA = √ 2L 2−
2 =√ (L/16) 4 2− 4/3 √ √ 2 = 12 + 4 6 ≈ 21.80 sec. 4/3 The time required for Alison to ﬁnish the race is less than Kevin; therefore Alison wins the √ √ race by 6 3 − 4 6 ≈ 0.594 seconds. EXERCISES 1.2: Solutions and Initial Value Problems, page 14 1. (a) Diﬀerentiating φ(x) yields φ (x) = 6x2 . Substitution φ and φ for y and y into the given equation, xy = 3y , gives x 6x2 = 3 2x3 , 3 ...
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Constant of integration, Alison

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