An object with a mass of 10.00kg is launched from the ground with an initial upward velocity of 95.00m/s near the surface of Earth. Consider a linear model of the air resistance, kv, with v being the variable velocity and k = 0.0365 kg/s?. Take the ground as the origin and upwards as positive direction. Which of the following best describes the DE relating to the problem? Let t be the time, g be the acceleration due to gravity. a. -10.00- 3.285 = 10.00 dv/dt b. -10.00g +0.0365v = 10.00 dv/dt c. -10.00g - 0.0365v = 10.00 dv/dt d. -10.00g – 3.285 = -10.00dv/dt Which of the following is closest to the general solution of the linear DE model, velocity as a function of time? Let C be a real constant, exp()=e0 a. v(t) = 1.346t + C b. v(t) = -2688+Cexp(t/274.0) c. v(t) = -2688+Cexp(-1/10.00) d. v(t) = -2688+Cexp(-1/274.0) %3D Which of the following is closest to the exact solution? a. t) = -2688+2783exp(-t/274.0) b. vt) = -2688+2783exp(-t/10.00) c. t) = -2688+2783exp(t/274.0) d. t) = 1.346t + 95.00 %3D

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter2: Equations And Inequalities
Section2.7: More On Inequalities
Problem 44E
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Can you please answer numbers 1 to 3? 

An object with a mass of 10.00kg is launched from the ground with an initial upward velocity of 95.00m/s near
the surface of Earth. Consider a linear model of the air resistance, kv, with v being the variable velocity and k =
0.0365 kg/s?. Take the ground as the origin and upwards as positive direction.
Which of the following best describes the DE relating to the problem? Let t be the time, g be the
acceleration due to gravity.
a. -10.00- 3.285 = 10.00 dv/dt
b. -10.00g +0.0365v = 10.00 dv/dt
c. -10.00g - 0.0365v = 10.00 đv/dt
d. -10.00g – 3.285 = -10.00dv/dt
Which of the following is closest to the general solution of the linear DE model, velocity as a function of
time? Let C be a real constant, exp()=e0
a. v(t) = 1.346t + C
b. v(t) = -2688+Cexp(t/274.0)
c. v(t) = -2688+Cexp(-1/10.00)
d. v(t) = -2688+Cexp(-1/274.0)
Which of the following is closest to the exact solution?
a. t) = -2688+2783exp(-t/274.0)
b. t) = -2688+2783exp(-t/10.00)
c. t) = -2688+2783exp(t/274.0)
d. t) = 1.346t + 95.00
Which of the following is closest to the time for the object to reach the maximum height?
a. 274.0s
b. 95.00s
c. 0.3473s
d. 9.517s
Which of the following is closest to the maximum height? hint: s = S vdt
а. 32.38m
b. 15100m
c. 449.4m
d. 16.40m
Transcribed Image Text:An object with a mass of 10.00kg is launched from the ground with an initial upward velocity of 95.00m/s near the surface of Earth. Consider a linear model of the air resistance, kv, with v being the variable velocity and k = 0.0365 kg/s?. Take the ground as the origin and upwards as positive direction. Which of the following best describes the DE relating to the problem? Let t be the time, g be the acceleration due to gravity. a. -10.00- 3.285 = 10.00 dv/dt b. -10.00g +0.0365v = 10.00 dv/dt c. -10.00g - 0.0365v = 10.00 đv/dt d. -10.00g – 3.285 = -10.00dv/dt Which of the following is closest to the general solution of the linear DE model, velocity as a function of time? Let C be a real constant, exp()=e0 a. v(t) = 1.346t + C b. v(t) = -2688+Cexp(t/274.0) c. v(t) = -2688+Cexp(-1/10.00) d. v(t) = -2688+Cexp(-1/274.0) Which of the following is closest to the exact solution? a. t) = -2688+2783exp(-t/274.0) b. t) = -2688+2783exp(-t/10.00) c. t) = -2688+2783exp(t/274.0) d. t) = 1.346t + 95.00 Which of the following is closest to the time for the object to reach the maximum height? a. 274.0s b. 95.00s c. 0.3473s d. 9.517s Which of the following is closest to the maximum height? hint: s = S vdt а. 32.38m b. 15100m c. 449.4m d. 16.40m
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ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage