Explanation Given: • Velocity: v ( t ) = 12 − 4 t m/s • Time interval [ 0 , 5 ] Key Concepts Used: Let s ( t ) be the position at time t of an object in linear motion. Then the object’s velocity is v ( t ) = s ' ( t ) and the integral of v ( t ) is equal to the net change in position or displacement over a time interval [ t 1 , t 2 ] . ∫ v ( t ) ⋅ d t t 1 t 2 = ∫ s ' ( t ) ⋅ d t t 1 t 2 = s ( t 2 ) − s ( t 1 ) . Where s ' ( t ) is the rate at which the quantity of displacement (or) position) of the object is changing. Distance and Displacement are different terms in a linear motion of any object. For an object in linear motion with velocity v ( t ) , Displacement during [ t 1 , t 2 ] = ∫ v ( t ) ⋅ d t t 1 t 2 . Distance travelled during [ t 1 , t 2 ] = ∫ | v ( t ) | ⋅ d t t 1 t 2 . Here in the problem it is given that a particle moving in a straight line with velocity v ( t ) = 12 − 4 t m/s.They ask us to find the displacement and distance of the particle over the time interval [ 0 , 5 ] . To find distance: If we factorize the velocity function v ( t ) = 12 − 4 t , we get t = 3 . So at t = 3 the given velocity becomes zero and before t = 3 it is positive and after t = 3 it is negative. Thereby the limit splits into [ 0 , 3 ] a n d [ 3 , 5 ] . Calculation: Description: Displacement= s ( t 2 ) − s ( t 1 ) = ∫ v ( t ) ⋅ d t t 1 t 2 . Distance travelled during [ t 1 , t 2 ] = ∫ | v ( t ) | ⋅ d t t 1 t 2 . Step 1 : Write the velocity of the particle which moves in a straight line. v ( t ) = 12 − 4 t Step 2 : Write the initial time and final time. t 1 = 0 t 2 = 5 Step 3 : Substitute these values in the given equation to find the displacement of the particle over the time interval [ 0 , 5 ] . s ( t 2 ) − s ( t 1 ) = ∫ v ( t ) ⋅ d t t 1 t 2 s ( 5 ) − s ( 0 ) = ∫ ( 12 − 4 t ) ⋅ d t 0 5 = [ 12 t − 4 ⋅ t 2 2 ] 0 5 = [ 12 t − 2 t 2 ] 0 5 Step 5 : Calculate the value of the above equation to get the solution of the given problem...

4th Edition

ISBN: 9781319434526

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 5.6, Problem 9E

To determine

• Displacement of particle which moves in a straight lie with given velocity over provided time interval

• Distance of particle which moves in a straight lie with given velocity over provided time interval

• A motion diagram

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Chapter 5.6, Problem 8E

Chapter 5.6, Problem 10E

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Consider the graphs of y = f(x) and y = g(x) in the given diagram y= f(x). y = g(x) Evaluate (f+g)(2) -5 Determine all for which g(x) < f(x) Determine all for which f(x) +3 = g(x)

I) For what value(s) of x does g(x) = -4? Separate multiple answers with commas as needed. J) Give the interval(s) of such that g(x) > 0. Use the union symbol between multiple intervals. K) Give the interval(s) of such that g(x) <0. Use the union symbol between multiple intervals.

need help on B

Chapter 5 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

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• Displacement of particle which moves in a straight lie with given velocity over provided time interval • Distance of particle which moves in a straight lie with given velocity over provided time interval • A motion diagram

Solution Summary: The author analyzes the motion diagram to determine the displacement and distance of a particle.