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 object by the parametric equations x = t ( 2 t + 1 ) 2 and y = 6 4 t + 3 . 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 ( t ( 2 t + 1 ) 2 ) = t ( 2 ( 2 t + 1 ) ( 2 ) ) + ( 2 t + 1 ) 2 ( 1 ) ( Since d d x ( u v ) = u d v d x + v d u d x ) = 8 t 2 + 4 t + 4 t 2 + 4 t + 1 = 12 t 2 + 8 t + 1 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 ( 12 t 2 + 8 t + 1 ) = 12 ( 2 ) t + 8 ( 1 ) t 0 ( Since d u n d x = n u n − 1 d u d x ) = 24 t + 8 When t = 0.5 , the x – component of acceleration becomes, a x | t = 0.5 = 24 ( 0.5 ) + 8 = 12 + 8 = 20 Obtain the y – component of velocity by computing the derivative, d y d t as shown below

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Chapter 24.3, Problem 9E

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The magnitude and direction of the acceleration when t=0.5, x=t(2t+1)2 and y=64t+3.

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Chapter 24.3, Problem 8E

Chapter 24.3, Problem 10E

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Chapter 24 Solutions

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The magnitude and direction of the acceleration when t = 0.5 , x = t ( 2 t + 1 ) 2 and y = 6 4 t + 3 .

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The magnitude and direction of the acceleration when t=0.5, x=t(2t+1)2 and y=64t+3.

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Chapter 24 Solutions