Explanation Given: The Initial velocity ( υ ) of the boat is 10 m/s . Radius of curvature ( ρ ) is 50 m . The rate of increase in the speed ( υ ˙ ) of the boat is ( 0.05 υ ) m/s 2 . Write the expression for the instantaneous acceleration. a = d υ d t d t = d υ a (1). Here, acceleration is a and rate of change of velocity with respect to time is ( d υ d t ) . Write the expression for normal acceleration. a n = υ 2 ρ (2). Here, normal acceleration is a n , constant velocity is υ and radius of curvature is ρ . Write the expression for the magnitude of acceleration. a = a t 2 + a n 2 (3). Here, acceleration is a , tangential acceleration is a t and normal acceleration is a n . Write the expression for the rate of increase in the speed ( υ ˙ ) of the boat.. υ ˙ = a t = ( 0.05 υ ) m/s 2 (4). The rate of increase in the speed υ ˙ is equal to acceleration a and tangential acceleration a t . Conclusion: Use the initial condition of 10 m /s for velocity ( υ ) and 0 s for time ( t ) . Substitute ( 0.05 υ ) m/s 2 for a in Equation (1). Integrate the Equation (1) at the limits 0 to t for t and 10 m /s to υ for υ . ∫ 0 t d t = ∫ 10 υ d υ ( 0.05 υ ) t = 20 ln υ 10 υ = ( 10 e t / 20 ) m /s (5)

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Author: HIBBELER

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Chapter 12, Problem 133P

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The magnitude of the particle’s acceleration (a) after 4 s .

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Chapter 12, Problem 132P

Chapter 12, Problem 134P

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The magnitude of the particle’s acceleration ( a ) after 4 s .

Solution Summary: The author explains the expression for the instantaneous acceleration and the rate of change of velocity with respect to time.

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The magnitude of the particle’s acceleration (a) after 4 s .

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