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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 7.4 - Problem 80e
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 7.4 - Problem 80e

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Skydiving A skydiver has a downward velocity given by

ISBN: 9780321570567 2

Solution for problem 80E Chapter 7.4

Calculus: Early Transcendentals | 1st Edition

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Problem 80E

Skydiving A skydiver has a downward velocity given by where t = 0 is the instant the skydiver starts Calling, g ? 9.8 m/s2 is the acceleration due to gravity, and V is the terminal velocity of the skydiver.a. Evaluate v(0)and and interpret these results.________________b. Graph the velocity function.________________c. Verify by integration that the position function is given by where s?(t)= v(t) and s(0) = 0.________________d. Graph the position function.(See the Guided Projects for more details on free fall and terminal velocity.)

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Problem 80ESkydiving A skydiver has a downward velocity given by where t = 0 is the instant the skydiver starts Calling, g 9.8 m/s2 is the acceleration due to gravity, and V is the terminal velocity of the skydiver.a. Evaluate v(0)and and interpret these results.________________b. Graph the velocity function.________________c. Verify by integration that the position function is given by where s(t)= v(t) and s(0) = 0.________________d. Graph the position function.(See the Guided Projects for more details on free fall and terminal velocity.)Answer; Step-1 of 4; Given velocity is : v(t) = V() , where t = 0 is the instant the sky driver starts calling , g 9.8m/is the acceleration due to gravity, and V is the terminal velocity of the skydriver.1. In this problem we need evaluate v(0) and Given : v(t) = V() = V() = V(0) = 0 , since Therefore , v(0) = 0 = V() = V () = V () = V (1) , since as t , then Therefore , = VStep- 2of 4; b) Graph of the velocity function is shown below: Step-3 of 4 : c) Given : Consider , v(t) = V() S(t) = = V , since , where C is constant. Take = p , then differentiate both sides with respect to t we get; (1) = That implies , dt = dp Therefore , the above integral becomes : s(t) = V = V = = ln|, since dx = ln |f(x)|+ C = ln| = [ln, since ln(ab) = ln(a) +ln(b) = [p, since = [, since p = = vt + ln( 1+ ) +C . Therefore , s(t) = vt + ln( 1+ ) +C . s(0) = v(0) + ln( 1+ ) +C . 0 = ln(2) +C , since = 1 C = - ln(2) Therefore , s(t) = vt + ln( 1+ ) - ln(2) . = vt + (ln( 1+ ) - ln(2)) = vt + (ln( 1+ )/2)) , since ln( a/b) = ln(a) - ln(b). Therefore , s(t) = vt + (ln( 1+ )/2)) .Step- 4 of 4; d) The graph of the related function is shown below : .

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