Explanation Given: A particle is moving with an accelaretion a ( t ) = t + t 2 . At t = 0 , s ( 0 ) = 0 , v ( 0 ) = 2 Where s ( t ) denotes the position of the particle after t seconds and v ( t ) denotes the velocity of the particle after t seconds. Concept used: Integration of a function can be defined by anti derivative of that function. For any function f ( x ) = b x + C , derivative of the function can be evaluated as f ′ ( x ) = b d d x ( x ) + d d x ( C ) = b ( 1 ) + 0 = b Therefore, antiderivative of b is given by ∫ b d x = b x + C Here, b is any constant and C is integrating constant. Formula used: Accelaration of a moving particle after time t is given by a ( t ) = d v d t = v ′ ( t ) Where v ( t ) is the velocity at time t such that v ( t ) = d s d t = s ′ ( t ) Where s ( t ) is the position of the partical after time t Linear property of inderfinite integral are as follows: ∫ [ a f ( x ) ± b g ( x ) ] d x = a ∫ f ( x ) d x ± b ∫ g ( x ) d x Formula to evaluate the integral of the form ∫ a x n d x is shown below. ∫ a x n d x = a ∫ x n d x = a x n + 1 n + 1 + C [ n ≠ − 1 ] Calculation: Now, consider the given acceleration. a ( t ) = t + t 2 Therefore, velocity after time t is calculated as shown below. ∫ a ( t ) d t = ∫ ( t + t 2 ) d t ∫ v ′ ( t ) d t = ∫ t d t + ∫ t 2 d t + C v ( t ) = [ t 1 2 + 1 1 2 + 1 ] + [ t 2 + 1 2 + 1 ] + C v ( t ) = [ t 3 2 3 2 ] + [ t 3 3 ] + C v ( t ) = 2 3 t 3 2 + t 3 3 + C Here, C is any integrating constant. Now, the initial velocity is 2. Therefore, at t = 0 , v ( 0 ) = 2

6th Edition

ISBN: 9781465208880

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 5.1, Problem 44PS

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To calculate:The position and velocity of the particle at t=4 .

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Chapter 5.1, Problem 43PS

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Chapter 5 Solutions

Calculus

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The position and velocity of the particle at t = 4 .

Solution Summary: The author explains how integration of a function can be defined by anti-derivative of b.

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Chapter 5 Solutions