Terminal velocity in a viscous medium

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An 11.00 kg object falling through a viscous medium reaches a terminal speed of 82.7 m/s and attains three-fourths of this speed in 11.7 seconds. The resistive force acting on the object is proportional to its velocity, described by R = -bv. The user seeks assistance in determining the distance traveled in the first 5.84 seconds and in formulating the correct equation of motion. The proposed equation for motion is incorrect due to the non-constant acceleration in this scenario, as the acceleration decreases over time due to the resistive force. The calculated distance of 483 m is questioned, indicating a misunderstanding of the motion dynamics in a viscous medium.
shiri
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A 11.00 kg object starting from rest falls through a viscous medium and experiences a resistive force R = -bv, where v is the velocity of the object. The object reaches one half its terminal speed in 5.84 s.

(a) Determine the terminal speed.
82.7 m/s

(b) At what time is the speed of the object three-fourths the terminal speed?
11.7 s

I couldn't figure out on this question. can someone help me out, i'll be appreciated
(c) How far has the object traveled in the first 5.84 s of motion?
 
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Can you write an equation of motion for the object?
 
Hootenanny said:
Can you write an equation of motion for the object?

was it v = [Initial velocity] + [(1/2)at^2]?
 
shiri said:
was it v = [Initial velocity] + [(1/2)at^2]?
That equation is only valid for constant acceleration, but is the acceleration constant in this case?
 
well what i got for this question is x=d=mgt/b + m^2g/b^2 * (exp(-bt/m) - 1)=483m
but I don't understand why this is a wrong answer so can you tell me why?
 
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