Test bank and solution of calculus (1)

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Test bank and solution of calculus (1)

2.1 SOLUTIONS 95CHAPTER TWOSolutions for Section 2.1--------------------------------------------------------------Exercises1.For t between 2 and 5, we

Test bank and solution of calculus (1)e haveAs ÍOO - 1352G5 ,Average velocity = — = —.— = —- knvnr.Ỡ 2•.»The average velocity on this part of the trip was 2G5/3 knưhr.2.The average velocit

y over a time period is the change in position divided by the change in time. Since the function r(t) gives the position of the particle, we find the Test bank and solution of calculus (1)

values of t(0) = -2 and z(4) = -6. Using these values, we find*(4)-x(0) -6-(-2)Average velocity = y = —--—= —-——- = -1 meters/sec.'1 — 03.The average

Test bank and solution of calculus (1)

velocity over a time period is the change in position divided by the change in time. Since the function x{t) gives the position of the particle, we fi

2.1 SOLUTIONS 95CHAPTER TWOSolutions for Section 2.1--------------------------------------------------------------Exercises1.For t between 2 and 5, we

Test bank and solution of calculus (1)stroms/sec.At 0 — 264.The average velocity over a lime period is the change in position divided by Ilk' change in time. Since the function «(t) gives

the distance of the particle from a point, we read off the graph that s(0) = 1 and s(3) = 4. Thus., . Astt) s(3) - .<0145017,Average velocity = -- = 1 Test bank and solution of calculus (1)

meter/sec.kJ’«5 ” u>55.The average velocity over a time period is the change in position divided by tlie change in time. Since Ilk* function s(f) giv

Test bank and solution of calculus (1)

es the distance of the particle from a point, we read off the graph that s(l) = 2 and s(3) = 6. Thus., MO s(3)-s(l)G —2 .._________,Average velocity =

2.1 SOLUTIONS 95CHAPTER TWOSolutions for Section 2.1--------------------------------------------------------------Exercises1.For t between 2 and 5, we

Test bank and solution of calculus (1)gives the distance of the particle from a point, we find the values of s(2) = c1 2 3 4 5 6 7 — 1 = 6.389 and .s(4) = c* — 1 = 53.598. Using these valu

es, we find, ... MO s(4)-«(2) 53.598-G.389Average velocity = —= —4—7^— =-----------------—--------— = 23.605 pm/sec.Ar I — 227.The average velocity ov Test bank and solution of calculus (1)

era time period is the change in the distance divided by the change in time. Since the function s(t) gives the distance of the particle from a point,

Test bank and solution of calculus (1)

we find the values of s(tr/3) = 4 + 3v/3/2 and s(7if/3) = 4 + 3v/5/2. Using these values, we findA^evdod.y == ■' 3ựã/2 - H + 3/Ì/2) =6} At 7ir/3 - tr/

2.1 SOLUTIONS 95CHAPTER TWOSolutions for Section 2.1--------------------------------------------------------------Exercises1.For t between 2 and 5, we

Test bank and solution of calculus (1)Let s = /(f).(i)We wish Io find the average velocity between I = 1 and t = 1.1. We haveAverage velocity = ZihlLzJl*) = 3-^~3 = 6.3 m/sec.

2.1 SOLUTIONS 95CHAPTER TWOSolutions for Section 2.1--------------------------------------------------------------Exercises1.For t between 2 and 5, we