Terminal Velocity given Time at which velocity is .5Vt

In summary, a 9.00 kg object falls through a viscous medium with a resistive force of R = -bv, where v is the velocity of the object. If the object reaches one-half its terminal speed in 5.93 s, the terminal speed can be determined using the equation v=mg/b(1-e^-bt/m). The speed after 5.93 s is mg/(2b) and from this condition, b can be found. The object has a terminal speed of 83.8403 m/s and reaches a speed of .75vt at a time of approximately 7.91 s. Lastly, after 5.93 s of motion, the object has traveled a distance of approximately
  • #1
xslc
6
0

Homework Statement



A 9.00 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. If the object reaches one-half its terminal speed in 5.93 s,
(a) determine the terminal speed.

Homework Equations



v=mg/b(1-e^-bt/m)

The Attempt at a Solution



Tried setting above equation equal to mg/2b and solving. It was wrong.
 
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  • #2
Show your work in detail. You were right, the speed after 5.93 s is
mg/(2b) From this condition, you can find b.

ehild
 
  • #3
ok, so:

v = mg/b(1-e^(-bt/m))=mg/(2b)
mg/b's cancel, so 1-e^(-5.93b/m)=.5
.5=e^-5.93b/9
9ln.5=-5.93b
-6.2383=-5.93b
b=1.052
vt = 88.2/b = 88.2/1.052=83.8403

and...i got it.

for some reason i was pluggin in 5.5 not 5.93, thanks for letting me know i was doing it right.
 
  • #4
Trust in yourself and check your calculations:smile:

ehild
 
  • #5
one more thing.

i got that, and i got part b, which asked for the time at which the speed is .75vt

i'm now stuck on part c, which says:

(c)How far has the object traveled in the first 5.93 s of motion?

I realize i need to integrate somehow, but I'm not sure exactly what to do...
 
  • #6
You know calculus, don't you?
Think of the definition of velocity: it is the time derivative of displacement. Integrating the velocity with respect to time from zero to a given moment will give the displacement. Try.

ehild
 
  • #7
got it. i was trying to do an indefinite integral and then plug in t, silly me, it was late last night haha.
 

Related to Terminal Velocity given Time at which velocity is .5Vt

1. What is terminal velocity?

Terminal velocity is the maximum velocity that an object can reach as it falls through a medium, such as air or water. This velocity is achieved when the drag force acting on the object is equal to the force of gravity pulling the object down.

2. How is terminal velocity calculated?

Terminal velocity is calculated using the formula Vt = (2mg)/ρAC, where Vt is terminal velocity, m is the mass of the falling object, g is the acceleration due to gravity, ρ is the density of the medium, A is the cross-sectional area of the object, and C is the drag coefficient.

3. What does it mean when the time at which velocity is .5Vt is given?

This means that the object has reached half of its terminal velocity. In other words, it has fallen through half of the distance where the drag force becomes equal to the force of gravity and it can no longer accelerate.

4. How does the time at which velocity is .5Vt relate to the shape and size of the object?

The time at which velocity is .5Vt is directly related to the shape and size of the object. Objects with a larger surface area or a lower drag coefficient will take longer to reach half of their terminal velocity compared to objects with a smaller surface area or a higher drag coefficient.

5. Can terminal velocity be reached in a vacuum?

No, terminal velocity cannot be reached in a vacuum because there is no medium for the object to fall through and experience drag. In a vacuum, objects will continue to accelerate until they reach the speed of light, if they are in a gravitational field.

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