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--------------------------------------------------------------Exercises1For t between 2 and 5, we

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

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 va Test bank and solution of calculus (1)

lues 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 — 03The average vel

Test bank and solution of calculus (1)

ocity 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 find

2.1 SOLUTIONS 95CHAPTER TWOSolutions for Section 2.1--------------------------------------------------------------Exercises1For t between 2 and 5, we

Test bank and solution of calculus (1)oms/sec.At 0 — 264The 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 met Test bank and solution of calculus (1)

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

Test bank and solution of calculus (1)

e 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--------------------------------------------------------------Exercises1For t between 2 and 5, we

Test bank and solution of calculus (1)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 values, we

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

e 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, we find

Test bank and solution of calculus (1)

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/32x'Tho

2.1 SOLUTIONS 95CHAPTER TWOSolutions for Section 2.1--------------------------------------------------------------Exercises1For t between 2 and 5, we

Test bank and solution of calculus (1) /(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--------------------------------------------------------------Exercises1For t between 2 and 5, we