Terminal velocity in a viscous medium

In summary, when an 11.00 kg object falls through a viscous medium with a resistive force of R = -bv, it reaches one half its terminal speed in 5.84 s. Using the equation v = (mg/b) * (1 - exp(-bt/m)), we can determine that the terminal speed is 82.7 m/s and the speed of the object is three-fourths the terminal speed at 11.7 s. The object travels approximately 483 m in the first 5.84 s of motion. However, the equation provided (x=d=mgt/b + m^2g/b^2 * (exp(-bt/m) - 1)) is not an accurate representation as it
  • #1
shiri
85
0
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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  • #2
Can you write an equation of motion for the object?
 
  • #3
Hootenanny said:
Can you write an equation of motion for the object?

was it v = [Initial velocity] + [(1/2)at^2]?
 
  • #4
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?
 
  • #5
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?
 

What is terminal velocity in a viscous medium?

Terminal velocity in a viscous medium refers to the maximum speed that an object can reach when falling through a fluid, such as air or water, that provides resistance or drag. This velocity occurs when the downward pull of gravity is balanced by the upward force of drag, leading to a constant speed.

How is terminal velocity in a viscous medium calculated?

The calculation of terminal velocity in a viscous medium takes into account the weight and surface area of the falling object, as well as the density and viscosity of the fluid. This can be calculated using the formula v = (2mg)/ (ρAC), where v is terminal velocity, m is the mass of the object, g is the acceleration due to gravity, ρ is the density of the fluid, A is the frontal area of the object, and C is the drag coefficient.

What factors affect terminal velocity in a viscous medium?

Terminal velocity in a viscous medium is affected by several factors, including the density and viscosity of the fluid, the surface area and shape of the falling object, and the force of gravity. Additionally, any changes in these factors during the fall, such as changes in air density due to altitude, can also impact the terminal velocity.

How does terminal velocity in a viscous medium differ from that in a non-viscous medium?

In a non-viscous medium, such as a vacuum, there is no drag force acting on the falling object, so it will continue to accelerate until it reaches its maximum velocity. In a viscous medium, the drag force increases with the speed of the object, eventually balancing out the downward pull of gravity and leading to a constant terminal velocity.

What are some real-life applications of terminal velocity in a viscous medium?

Terminal velocity in a viscous medium is important for understanding the behavior of objects falling through fluids, such as parachutes, skydivers, and raindrops. It is also relevant in industries like aviation, where the drag force acting on airplanes can impact their speed and fuel efficiency. In medicine, understanding terminal velocity in a viscous medium can help with the development of drug delivery systems and the study of blood flow in capillaries.

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