Application of Newton's second law of motion EF = ma to a 0.5-kg object thrown straight up with an initial velocity of 4 m/s and adopting the sign convention that down is positive results in the following equation: mg - 2v = ma. What is the velocity v of the object as a function of t? %3D

Calculus: Early Transcendentals
8th Edition
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Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Application of Newton's second law of motion
EF = ma to a 0.5-kg object thrown straight up
with an initial velocity of 4 m/s and adopting
the sign convention that down is positive
results in the following equation: mg - 2v = ma.
What is the velocity v of the object as a
function of t?
Select the correct response:
v = 2.45 - 6.45 exp (-4t)
O v = 2.45 - 1.55 exp (-4t)
O v = 2.45 + 1.55 exp (-4t)
V = 2.45 + 6.45 exp (4t)
Transcribed Image Text:Application of Newton's second law of motion EF = ma to a 0.5-kg object thrown straight up with an initial velocity of 4 m/s and adopting the sign convention that down is positive results in the following equation: mg - 2v = ma. What is the velocity v of the object as a function of t? Select the correct response: v = 2.45 - 6.45 exp (-4t) O v = 2.45 - 1.55 exp (-4t) O v = 2.45 + 1.55 exp (-4t) V = 2.45 + 6.45 exp (4t)
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