Berkeley Review Physics chapter 2: terminal velocity.

[ibeatupnerds]

I don't want to get in trouble for copyright so i'll adjust the question. Say a ball falls from rest, the question asks you to compare velocity and time.

we can eliminate A and C, since it shows a speed greater than Vt. The book picks D because according to the book the speed doesn't "abruptly" change (basically they are saying in nature things naturally change so the curved graph in D is correct. they are basically focusing on the short time interval very close to terminal velocity). I picked B because the slope of velocity vs. time shows acceleration. And ,of course, acceleration is a constant g, which is justified by the straight line prior terminal velocity (vt). Any comments?

Or is this just not a good question and the correct answer should look like B at the start but slowly transition into Vt like in D.

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[exquisitemelody]

(sorry, I'm editing my answer...hope you didn't see what was there before)

Because we're talking about terminal velocity, that means that we're not ignoring air resistance. Terminal velocity occurs when the acceleration due to gravity equals the drag force. At the beginning, the force of gravity is more dominant than the drag force. However, as the object falls, the drag force becomes larger and larger, which causes the acceleration due to gravity to become smaller. When terminal velocity is reached, the force of gravity of the object equals the drag force, which means that there's no acceleration.

You can only assume constant acceleration due to gravity when there are no external forces. In this case, there is an external force - drag force.

http://en.wikipedia.org/wiki/Terminal_velocity

 

[whiteshadodw]

basically as speed increases, drag increases as well, and eventually the force of gravity and force due to drag cancel out. this means there will be some maximum velocity the ball can reach.

so we can narrow it down to B and D. in B, the change in acceleration is not gradual, its a very abrupt change, as if gravity just disappeared. in D we can see a more gradual change in acceleration as it approaches zero.

 

[BerkReviewTeach]

(sorry, I'm editing my answer...hope you didn't see what was there before)

Because we're talking about terminal velocity, that means that we're not ignoring air resistance. Terminal velocity occurs when the acceleration due to gravity equals the drag force. At the beginning, the force of gravity is more dominant than the drag force. However, as the object falls, the drag force becomes larger and larger, which causes the acceleration due to gravity to become smaller. When terminal velocity is reached, the force of gravity of the object equals the drag force, which means that there's no acceleration.

You can only assume constant acceleration due to gravity when there are no external forces. In this case, there is an external force - drag force.

EXCELLENT explanation!!!

The acceleration gradually goes from g to 0 as the objects speeds up, so the tangent to the graph must flatten out gradually until it reaches a slope of zero. Choice B fits that description while choice D does not.

 

[cge0]

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[MD Odyssey]

The BR chapter this problem is from covers forces, so looking at the problem in that context might be helpful. There are two forces on the falling body - the force due to gravity and the upward force due to friction, which is generally proportional to the square of the velocity. Finding the net force and setting it equal to ma leads to an equation of motion which can be solved using calculus.

An alternate and more intuitive approach is to recognize that, since the frictional force is proportional to the square of the velocity, the initial acceleration is simply the acceleration due to gravity. So, initially, the acceleration of the body is a constant, as you mentioned, thus the velocity is linear. However, as the velocity increases, the frictional force increases as well, so the net acceleration is somewhat less than g. Over time, this continues until, at some point, the system reaches equilibrium, the net force on the object is zero, and the particle reaches terminal velocity.

Now, look at the velocity of the particle in graph B. The initial velocity is linear, as we would expect, but it abruptly changes once it reaches terminal velocity. This would mean that the acceleration instantaneously went from being a non-zero constant to zero, which is clearly incorrect. In general, natural phenomena do not change instantaneously, and we should immediately become suspicious of functions that do.

Mathematically, acceleration is defined as the time derivative of velocity, which we know to exist at every point t > 0. But, the function in graph B has a cusp and therefore fails to be differentiable at that point and thus does not exist there. What I mentioned earlier is a general principle that is often useful in answering MCAT style questions - graphs which imply an instantaneous change in a function's value are rarely correct because they contradict the mean value theorem.

Personally, the latter explanation appeals to me, but if you haven't had calculus, then ignore it and just focus on the other parts.