7_pdfsam_math 54 differential equation solutions odd

7_pdfsam_math 54 differential equation solutions odd -...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online