Explanation Procedure used: To find the acceleration of an object whose coordinates are given in parametric form: 1 Determine the x – component of acceleration a x by finding the derivative d v x d t where v x is the x – component of velocity. 2. Determine the y – component of acceleration a y by finding the derivative d v y d t where v y is the y – component of velocity. 3. Compute the magnitude of acceleration by using the formula, a = a x 2 + a y 2 and the direction of acceleration by the formula, tan θ = a y a x . Calculation: Given that the x and y – coordinates of position of the object by the parametric equations x = 0.8 t 5 2 and y = 1 − t 2 . Obtain the x – component of velocity by computing the derivative, d x d t as shown below. v x = d x d t = d d t ( 0.8 t 5 2 ) = 0.8 ( 5 2 ) t 5 2 − 1 ( Since d u n d x = n u n − 1 d u d x ) = 4 2 t 3 2 = 2 t 3 2 Now obtain the x – component of acceleration by computing the derivative, d v x d t as shown below. a x = d v x d t = d d t ( 2 t 3 2 ) = 2 ( 3 2 ) t 3 2 − 1 ( Since d u n d x = n u n − 1 d u d x ) = 3 t 1 2 When t = 2 , the x – component of acceleration becomes, a x | t = 2 = 3 ( 2 ) 1 2 = 3 2 Obtain the y – component of velocity by computing the derivative, d y d t as shown below

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Differential Equations

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Chapter 24.3, Problem 1PE

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The magnitude and direction of the acceleration when t=2, x=0.8t52 and y=1−t2.

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Chapter 24.2, Problem 30E

Chapter 24.3, Problem 1E

Chapter 24 Solutions

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Ch. 24.1 - Prob. 9E Ch. 24.1 - Prob. 10E Ch. 24.1 - Prob. 11E Ch. 24.1 - Prob. 12E Ch. 24.1 - In Exercises 11–14, find the equations of the...Ch. 24.1 - Prob. 14E Ch. 24.1 - Prob. 15E Ch. 24.1 - Prob. 16E Ch. 24.1 - Prob. 17E Ch. 24.1 - Prob. 18E Ch. 24.1 - Prob. 19E Ch. 24.1 - Prob. 20E Ch. 24.1 - Prob. 21E Ch. 24.1 - Where does the normal line to the parabola y = x —...Ch. 24.1 - Prob. 23E Ch. 24.1 - Prob. 24E Ch. 24.1 - A certain suspension cable with supports on the...Ch. 24.1 - Prob. 26E Ch. 24.1 - Prob. 27E Ch. 24.1 - Prob. 28E Ch. 24.1 - Prob. 29E Ch. 24.1 - Prob. 30E Ch. 24.2 - In Example 1, let x1 = 0.3, and find x2. 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Prob. 48RE Ch. 24 - Prob. 49RE Ch. 24 - Prob. 50RE Ch. 24 - Prob. 51RE Ch. 24 - Prob. 52RE Ch. 24 - In Exercises 49–94, solve the given problems. 53....Ch. 24 - Prob. 54RE Ch. 24 - Prob. 55RE Ch. 24 - Prob. 56RE Ch. 24 - The deflection y (in m) of a beam at a horizontal...Ch. 24 - Prob. 58RE Ch. 24 - Prob. 59RE Ch. 24 - Prob. 60RE Ch. 24 - Prob. 61RE Ch. 24 - Prob. 62RE Ch. 24 - In Fig. 24.75, the tension T supports the 40.0-N...Ch. 24 - Prob. 64RE Ch. 24 - Prob. 65RE Ch. 24 - Prob. 66RE Ch. 24 - An analysis of the power output P (in kW/m3) of a...Ch. 24 - The altitude h (in ft) of a certain rocket as a...Ch. 24 - Prob. 69RE Ch. 24 - Prob. 70RE Ch. 24 - Prob. 71RE Ch. 24 - Prob. 72RE Ch. 24 - Prob. 73RE Ch. 24 - A special insulation strip is to be sealed...Ch. 24 - Prob. 75RE Ch. 24 - Prob. 76RE Ch. 24 - Prob. 77RE Ch. 24 - Prob. 78RE Ch. 24 - Prob. 79RE Ch. 24 - Prob. 80RE Ch. 24 - Prob. 81RE Ch. 24 - Prob. 82RE Ch. 24 - Prob. 83RE Ch. 24 - Prob. 84RE Ch. 24 - Prob. 85RE Ch. 24 - Prob. 86RE Ch. 24 - 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The magnitude and direction of the acceleration when t = 2 , x = 0.8 t 5 2 and y = 1 − t 2 .

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The magnitude and direction of the acceleration when t=2, x=0.8t52 and y=1−t2.

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Chapter 24 Solutions