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

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