a. Answer t = 16 / 49 = 4 / 7 seconds . V a v e r a g e = ( 160 − 0 ) / ( 4 / 7 ) = 280 c m / sec . Explanation Given information: Suppose two balls are falling from rest at a certain height in centimeters above the ground. Use the equation s = 490 t 2 to answer the following questions. Calculation: s = 490 t 2 . putting s = 160. 160 = 490 t 2 t = 16 / 49 = 4 / 7 seconds . Slope formula V a v e r a g e = ( s ( b ) − s ( a ) ) / ( b − a ) . V a v e r a g e = ( s ( 4 / 7 ) − s ( 0 ) ) / ( ( 4 / 7 ) − 0 ) . V a v e r a g e = ( 160 − 0 ) / ( 4 / 7 ) = 280 c m / sec . t = 16 / 49 = 4 / 7 seconds . V a v e r a g e = ( 160 − 0 ) / ( 4 / 7 ) = 280 c m / sec . b. To determine To find: The velocity and the acceleration. Answer The velocity at t=4/7 is 560 cm/sec; the acceleration at t=4/7 is 980 c m 2 / s e c . Explanation Given information: Suppose two balls are falling from rest at a certain height in centimeters above the ground. Use the equation s = 490 t 2 to answer the following questions. Calculation: s ( t ) = 490 t 2 . s ' ( t ) = 980 t . s n ( t ) = 980. s ' ( 4 / 7 ) = 560. s n ( 4 / 7 ) = 980. The velocity at t=4/7 is 560 cm/sec; the acceleration at t=4/7 is 980 c m 2 / s e c .

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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Question

Chapter 3, Problem 72RE

To determine

To find: The time taken by balls to fall the first 160cm and the average velocity for the period.

Expert Solution

Answer to Problem 72RE

t=16/49=4/7 seconds.

Vaverage=(160−0)/(4/7)=280 cm/sec.

Explanation of Solution

Given information: Suppose two balls are falling from rest at a certain height in centimeters above the ground. Use the equation s=490t2 to answer the following questions.

Calculation:

s=490t2.putting s=160.160=490t2t=16/49=4/7 seconds.Slope formulaVaverage=(s(b)−s(a))/(b−a).Vaverage=(s(4/7)−s(0))/((4/7)−0).Vaverage=(160−0)/(4/7)=280 cm/sec.

t=16/49=4/7 seconds.

Vaverage=(160−0)/(4/7)=280 cm/sec.

To determine

To find: The velocity and the acceleration.

Expert Solution

Answer to Problem 72RE

The velocity at t=4/7 is 560 cm/sec; the acceleration at t=4/7 is 980 cm2/sec.

Explanation of Solution

Given information: Suppose two balls are falling from rest at a certain height in centimeters above the ground. Use the equation s=490t2 to answer the following questions.

Calculation:

s(t)=490t2.s'(t)=980t.sn(t)=980.s'(4/7)=560.sn(4/7)=980.

The velocity at t=4/7 is 560 cm/sec; the acceleration at t=4/7 is 980 cm2/sec.

Chapter 3, Problem 71RE

Chapter 3, Problem 73RE

Chapter 3 Solutions

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

Ch. 3.1 - Prob. 1QR Ch. 3.1 - Prob. 2QR Ch. 3.1 - Prob. 3QR Ch. 3.1 - Prob. 4QR Ch. 3.1 - Prob. 5QR Ch. 3.1 - Prob. 6QR Ch. 3.1 - Prob. 7QR Ch. 3.1 - Prob. 8QR Ch. 3.1 - Prob. 9QR Ch. 3.1 - Prob. 10QR

Ch. 3.1 - Prob. 1E Ch. 3.1 - Prob. 2E Ch. 3.1 - Prob. 3E Ch. 3.1 - Prob. 4E Ch. 3.1 - Prob. 5E Ch. 3.1 - Prob. 6E Ch. 3.1 - Prob. 7E Ch. 3.1 - Prob. 8E Ch. 3.1 - Prob. 9E Ch. 3.1 - Prob. 10E Ch. 3.1 - Prob. 11E Ch. 3.1 - Prob. 12E Ch. 3.1 - Prob. 13E Ch. 3.1 - Prob. 14E Ch. 3.1 - Prob. 15E Ch. 3.1 - Prob. 16E Ch. 3.1 - Prob. 17E Ch. 3.1 - Prob. 18E Ch. 3.1 - Prob. 19E Ch. 3.1 - Prob. 20E Ch. 3.1 - Prob. 21E Ch. 3.1 - Prob. 22E Ch. 3.1 - Prob. 23E Ch. 3.1 - Prob. 24E Ch. 3.1 - Prob. 25E Ch. 3.1 - Prob. 26E Ch. 3.1 - Prob. 27E Ch. 3.1 - Prob. 28E Ch. 3.1 - Prob. 29E Ch. 3.1 - Prob. 30E Ch. 3.1 - Prob. 31E Ch. 3.1 - Prob. 32E Ch. 3.1 - Prob. 33E Ch. 3.1 - Prob. 34E Ch. 3.1 - Prob. 35E Ch. 3.1 - Prob. 36E Ch. 3.1 - Prob. 37E Ch. 3.1 - Prob. 38E Ch. 3.1 - Prob. 39E Ch. 3.1 - Prob. 40E Ch. 3.1 - Prob. 41E Ch. 3.1 - Prob. 42E Ch. 3.1 - Prob. 43E Ch. 3.1 - Prob. 44E Ch. 3.1 - Prob. 45E Ch. 3.2 - Prob. 1QR Ch. 3.2 - Prob. 2QR Ch. 3.2 - Prob. 3QR Ch. 3.2 - Prob. 4QR Ch. 3.2 - Prob. 5QR Ch. 3.2 - 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Prob. 57RE Ch. 3 - Prob. 58RE Ch. 3 - Prob. 59RE Ch. 3 - Prob. 60RE Ch. 3 - Prob. 61RE Ch. 3 - Prob. 62RE Ch. 3 - Prob. 63RE Ch. 3 - Prob. 64RE Ch. 3 - Prob. 65RE Ch. 3 - Prob. 66RE Ch. 3 - Prob. 67RE Ch. 3 - Prob. 68RE Ch. 3 - Prob. 69RE Ch. 3 - Prob. 70RE Ch. 3 - Prob. 71RE Ch. 3 - Prob. 72RE Ch. 3 - Prob. 73RE Ch. 3 - Prob. 74RE Ch. 3 - Prob. 75RE Ch. 3 - Prob. 76RE Ch. 3 - Prob. 77RE Ch. 3 - Prob. 78RE Ch. 3 - Prob. 79RE Ch. 3 - Prob. 80RE Ch. 3 - Prob. 81RE Ch. 3 - Prob. 82RE Ch. 3 - Prob. 83RE

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