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 = 5 t 2 t + 1 and y = 0.1 ( t 2 + t ) . 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 ( 5 t 2 t + 1 ) = ( 2 t + 1 ) ( 5 ) − 5 t ( 2 ) ( 2 t + 1 ) 2 ( Since d d x ( u v ) = v ( d u d x ) − u ( d v d x ) v 2 ) = 10 t + 5 − 10 t ( 2 t + 1 ) 2 = 5 ( 2 t + 1 ) 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 ( 5 ( 2 t + 1 ) − 2 ) = 5 ( − 2 ) ( 2 t + 1 ) − 3 ( 2 ) ( Since d u n d x = n u n − 1 d u d x ) = − 20 ( 2 t + 1 ) 3 When t = 2 , the x – component of acceleration becomes, a x | t = 2 = − 20 ( 2 ( 2 ) + 1 ) 3 = − 20 ( 5 ) 3 = − 20 125 = − 0

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Author: Allyn Washington, Richard Evans

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Chapter 24.3, Problem 8E

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The magnitude and direction of the acceleration when t=2, x=5t2t+1 and y=0.1(t2+t).

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Chapter 24.3, Problem 7E

Chapter 24.3, Problem 9E

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Chapter 24 Solutions

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The magnitude and direction of the acceleration when t = 2 , x = 5 t 2 t + 1 and y = 0.1 ( t 2 + t ) .

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The magnitude and direction of the acceleration when t=2, x=5t2t+1 and y=0.1(t2+t).

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Chapter 24 Solutions