How far does a bubble in a glass of beer travel in its first and last second?

[Bogus_Roads]

Homework Statement

A bubble in a glass of beer releases from rest at the bottom of the glass and rises at acceleration, a, to the surface in t seconds. How much farther does the bubble travel in its last second than in its first second?

A) at
B)(t-1)a
C)(t+1)a
D)(1/2)at

Homework Equations

y=vt+(1/2)at^2

The Attempt at a Solution

1st sec: (1/2)at^2=(1/2)a
2nd sec: Vt+(1/2)at^2=V+(1/2)a=at+(1/2)a
2nd-1st=at+(1/2)a-(1/2)a=at

I think it's A but the book says B bc 2nd sec should be Vt-(1/2)at^2, but why would the acceleration be negative?

Thanks

 

[gneill]

What makes you think that the acceleration would be negative?

By the way, the problem statement is a bit sloppy in its expressions for the answers; the units don't quite work out to length. If t is in seconds, then the "1" constants should be in seconds, too, and all of the expressions should be multiplied by another unit of seconds (for example, in A), a*t gives units of velocity, not distance). Alternatively, t could be a unit-less count of seconds and each expression should then be multiplied by a factor of s² to yield distance measures.

 

[Bogus_Roads]

Yeah, I don't think it would be negative...the book says the distance of the last second=at-(1/2)a, which implies that acceleration is negative, which makes no sense. Mind you, it's an MCAT practice problem...

 

[Bogus_Roads]

Good point about units, though.

 

[soothsayer]

I don't see how it is implied that the acceleration is negative, the answer b) (t-1)a should be positive as long as t is greater than 1 second (which seems to be implied) so the bubble travels faster at the top than it does at the bottom, thus acceleration is positive (that is, if you define positive acceleration to be toward the top of the beer glass, of course).

That being said, it's hard for me to come up with an answer or explanation of the given answer, because I'm quite confused by the units as well.

 

[gneill]

Yeah, I don't think it would be negative...the book says the distance of the last second=at-(1/2)a, which implies that acceleration is negative, which makes no sense. Mind you, it's an MCAT practice problem...

a(t - 1/2) does not make acceleration negative. And so long as t > 1/2, the distance traveled is positive, too, and it should be for the "last second" of the travel.

 

[Bogus_Roads]

If x=Vt-(1/2)at^2, that implies that the acceleration is in the opposite direction of the velocity...that it's negative. The only way you could get B if this is true of the distance traveled in the last second. Otherwise, Vt+(1/2)at^2 (which implies that acceleration is in the same direction as the velocity), would give the answer A. The book insists it's B, so x=Vt-(1/2)at^2, so acceleration must be negative.

 

[Bogus_Roads]

I don't see that you can get from the givens to a(t-1/2) without saying acceleration (ONLY IN THE LAST SECOND) is negative.

 

[gneill]

Instead of trying to work with velocities, consider only the total distance traveled from a standing start at the bottom of the glass. What's the distance after 1 second? After t seconds? After t-1 seconds? The distance traveled during the last second is...?

 

[Bogus_Roads]

Distance depends on velocity..

 

[Bogus_Roads]

D1=1/2at^2.
D2=Vt+1/2at^2.
I don't know any other way to say it. The book, however, distinguishes d=Vnaughtt +1/2at^2 from d=vt-1/2at^2, which I don't understand.

 

[gneill]

D1=1/2at^2.
D2=Vt+1/2at^2.
I don't know any other way to say it. The book, however, distinguishes d=Vnaughtt +1/2at^2 from d=vt-1/2at^2, which I don't understand.

Suppose t is a unit-less count of seconds. Then the various distances involved are:

D₁ = (1/2)*a*(1s)²

D_t = (1/2)*a*t²s²

D_t-1 = (1/2)*a*(t - 1)²s²

The distance traversed during the final second is...?

 

[phever]

at t=(t-1) v=u+at hence v = a(t-1)

s = ut + 1/2 at^2
s = at(t-1) + 1/2a(t-1)^2
s will be positive

 

[Bogus_Roads]

The question asks for the difference between the distances covered in the first and last seconds, each 1 second long intervals, so t=1 for each. The total time appears in the answer just because it happens to be left over after one term is subtracted from the other.

"D₁ = (1/2)*a*(1s)2"

This is true...

"D_t = (1/2)*a*t2s2"

This is not; it will travel farther than this because of it's initial velocity at time=t.

"D_t-1 = (1/2)*a*(t - 1)2s2"

If you were to say t=5, then t-1=4, and D would designate the distance traveled in 4 seconds (or 4 counts)...

So, D_t - D_t-1 is the distance traveled in the last second.

Then, subtracting D₁ would give us the difference.

I'm a bit confused. How does that clear things up?

 

[gneill]

The more complete version of the displacement equation goes like:

d(t) = d0 + v0*t + (1/2)*a*t²

for total elapsed time t. If you start from rest at the origin of your chosen frame of reference, then d0 = 0, v0 = 0, and

d(t) = (1/2)*a*t²


That is, if you are computing the total distance starting from rest (v0 = 0) for an object with constant acceleration a traveling for time t, then that distance is given by

d(t) = (1/2)*a*t²

Note that you do not required the velocity at time t-1, or t-2, or t-1/1,000,000 in order to calculate d. You also did not need the distance already traveled at any of those intermediate times.

I'm not sure why you are hung up with the velocity at time t-1.

 

[Bogus_Roads]

"That is, if you are computing the total distance starting from rest (v0 = 0)", it's NOT starting from rest during the last second! I'm hung up on velocity because it gives the rectangular area under the velocity graph (displacement). Something traveling initially at 40 m/s with an acceleration of 5 m/s is not going to go 1/2(5 m/s)(2 s)^2 in 2 seconds, that would only be the distance traveled due to the increase in velocity (the trianglular area under the velocity graph)...it would go an additional (40 m/s)(2 s), right?

 

[Bogus_Roads]

Aha! V_t gives at, which gives the rectangular area it would travel if its velocity had been its max velocity during the last second, so (1/2)*a*t^2 must be subtracted to give the proper displacement. Otherwise, it should be v_t-1=a*(t-1). Does this make sense to anyone but me?

 

[gneill]

"That is, if you are computing the total distance starting from rest (v0 = 0)", it's NOT starting from rest during the last second! I'm hung up on velocity because it gives the rectangular area under the velocity graph (displacement). Something traveling initially at 40 m/s with an acceleration of 5 m/s is not going to go 1/2(5 m/s)(2 s)^2 in 2 seconds, that would only be the distance traveled due to the increase in velocity (the trianglular area under the velocity graph)...it would go an additional (40 m/s)(2 s), right?

It doesn't matter; if the total distance from time zero to time t is d2, and the total distance from time zero to time t-1 is d1, then the distance from time t-1 to t is d2 - d1. The (1/2)at² formula automatically takes care of the changing velocity all along the trajectory. Remember, the bubble doesn't stop and start again anywhere along its path.

 

[Bogus_Roads]

The (1/2)at² formula automatically takes care of the changing velocity all along the trajectory.

Do you agree that x=v₀t+1/2at² $$\neq$$ 1/2at²?

If so, how could 1/2at² describe the distance traveled?

 

[gneill]

Do you agree that x=v₀t+1/2at² $$\neq$$ 1/2at²?

If so, how could 1/2at² describe the distance traveled?

What's v₀ at t = 0? Each distance is being measured from t = 0 to some later value of t.

If you measure a distance from the origin to some point A, then the distance from the origin to some point B, what's the distance between points A and B?

 

[SammyS]

Aha! V_t gives at, which gives the rectangular area it would travel if its velocity had been its max velocity during the last second, so (1/2)*a*t^2 must be subtracted to give the proper displacement. Otherwise, it should be V_t-1=a*(t-1). Does this make sense to anyone but me?

I read this thread a few times in the last day and refrained from entering the discussion. Frankly, I have found it difficult to understand Bogus_Roads's argument as to why the acceleration must be negative during the last second if one accepts the book's answer.

I have to say that gneill's analysis is perfectly valid. It's true that it doesn't include velocity, but it doesn't need to. It also shows that the book answer can be obtained without resorting to using a negative acceleration.

Aha! V_t gives at, which gives the rectangular area it would travel if its velocity had been its max velocity during the last second

Finally I think I see where you're coming from. Using subscripts to indicate which velocity you're talking about sure helps!

Yes, the velocity, V_t, at time t is given by V_t= a*t. And you're correct to say that the distance the bubble would travel if it had this velocity for one second would be V_t*(1s)=V_t=a*t. But since the bubble has been accelerating (positively), its velocity was less than V_t during the last second of time, except right at the final time, t. The displacement during the final second is indeed
x_t - x_t-1 = V_t*(1s) - (1/2)*a*(1s)² = a*t*(1s) - (1/2)*a*(1s)² = a*t - (1/2)*a. This does NOT mean that the acceleration is negative. It means that at an earlier time the velocity was less that it was at the final time.

You can get the same answer by looking at the velocity, V_t-1, at time, t-1, i.e. 1 second prior to the bubble reaching the surface.
V_t-1 = a*(t-1) = a*t -a*1 . Of course if we multiply this by the time interval of 1 second, the resulting displacement will be too small by the amount, (1/2)*a*(1s)². The displacement during the final second is:
x_t - x_t-1 = V_t-1*(1s) + (1/2)*a*(1s)² = a*(t-1)*(1s) + (1/2)*a*(1s)² = a*t - a + (1/2)*a = a*t - (1/2)*a , which was obtained without any negative acceleration, although the last expression may appear as if a < 0.

By the way, using your reasoning, you could just as easily argue that the acceleration during the first second is negative.
At time t=1, V₁ = a*(1s) = a, and V₁*(1s) = a. But the displacement during the first second is only
x₁ - x₀ = (1/2)*a*(1s)² = (1/2)*a, so you need to subtract (1/2)*a. I.e. you can get the displacement during the first second by
x₁ - x₀ = V₁*(1s) - (1/2)*a*(1s)² = a - (1/2)*a = (1/2)*a, which may look as if a negative acceleration were used.

 

[phever]

negative acceleration could come from the viscous force by the liquid on the bubble?

 

[Bogus_Roads]

Awesome, that's exactly what I was trying to say. I was, before a day or so ago, unfamiliar with the idea that you could use V_f in the spot that I thought was reserved for V_o in the distance equation (and then obviously the (1/2)at² is subtracted, which made me think acceleration was negative). Thanks for your comment, SammyS, you said what I was trying to say in the previous post much better than I did.

Gneill, after playing around with graphs for a while, I finally understand that you're absolutely correct, and thanks for staying with me this long. It took me a while to realize that d_t was the entire displacement (the triangle with base t), d_t-1 was everything traveled in time t-1 (the triangle with base t-1), and taking the difference gives V_t-1t+1/2at². I kind of assumed that because and then subtracting 1/2a gave the difference in distance between the last and first seconds, which was, ironically, a(t-1)(1), or V_t-1. :)

Thanks again for the help.

 

[SammyS]

"That is, if you are computing the total distance starting from rest (v0 = 0)", it's NOT starting from rest during the last second! I'm hung up on velocity because it gives the rectangular area under the velocity graph (displacement). Something traveling initially at 40 m/s with an acceleration of 5 m/s is not going to go 1/2(5 m/s)(2 s)^2 in 2 seconds, that would only be the distance traveled due to the increase in velocity (the trianglular area under the velocity graph)...it would go an additional (40 m/s)(2 s), right?

Of course, but something traveling initially at 40 m/s with an acceleration of 5 m/s will travel (1/2)*(5 m/s²)*(2 s)² farther than it would if it had a constant velocity of 40 m/s for the 2 seconds.

Let's look at the situation described in the above quote. In the 2 s time interval, the object is displaced

Δx = (40 m/s)*(2 s) + (½)*(5 m/s²)*(2 s)² = 80 m + 10 m = 90 m .

I think we can all agree with that.

Analyzing a velocity versus time graph, the displacement is represented by the area under the velocity graph corresponding to these two seconds. The shape of this is a trapezoid. The area may be calculated in a number of ways. The 2 seconds is the altitude of the trapezoid, the two bases are 40 m/s and 50 m/s.
Area = (½)*(40 m/s + 50 m/s)*(2 s) = 90 m. Note: For constant acceleration, (½)*(40 m/s + 50 m/s) = 45 m/s is the average velocity for the 2 second time period. You can also find the area by taking the area of the (40 m/s)×(2 s) rectangle and add to that the area of the triangle, which represents the "extra" displacement caused by the acceleration. This triangle has a base of (2 s) and an altitude of (5 m/s²)*(2 s). The altitude represents the amount that the velocity increased.
Area of rectangle + Area of triangle = (40 m/s)*(2 s) + (½)*(2 s)*(5 m/s²)*(2 s) = 80 m + 10 m = 90 m.

The area can also be obtained by taking the (50 m/s)×(2 s) rectangle, then subtracting the area of the rectangle that happens to be congruent to the above triangle and shares its hypotenuse.
Area of rectangle − Area of triangle = (50 m/s)*(2 s) − (½)*(2 s)*(5 m/s²)*(2 s) = 100 m − 10 m = 90 m. Subtracting the area of the rectangle does not imply that the acceleration is negative. It reflects the realization that the velocity at the beginning of the 2 second time interval was less than the velocity at the end and if we used the velocity at the end of the interval multiplied by the time of 2 seconds the displacement would be too large.

The kinematic equation which corresponds to the last way we found the area is:
Δx = (50 m/s)*(2 s) − (½)*(5 m/s²)*(2 s)² = 100 m − 10 m = 90 m. That's the one which seems to me is giving you the trouble, because it seems to indicate that the acceleration is negative. What it really shows is that something with an acceleration of 5 m/s having a final velocity of 50 m/s will travel (1/2)*(5 m/s²)*(2 s)² less far in the previous 2 seconds than it would have if it had a constant velocity of 50 m/s for the 2 seconds.

As a final note: Suppose the above object had previously been at rest, then was accelerated at a constant acceleration of 5 m/s². It would have taken the object 8 seconds to reach a velocity of 40 m/s. Using the "total displacement analysis", the displacement from the starting position to the 40 m/s position would be:
Δx_8s = (½)*(5 m/s²)*(8 s)² = 160 m.
The displacement from the starting position to the 50 m/s position would be:
Δx_10s = (½)*(5 m/s²)*(10 s)² = 250 m.
The difference between these two is: Δx_10s − Δx_8s = 250 m − 160 m = 90 m.

I will repeat, there is absolutely nothing wrong with gneil's analysis. If you know the location of the bubble at times: T=0, T=1, T=t-1, and T=t, then you can certainly compare how far the bubble travels in the first second and in the last second.

 

[Bogus_Roads]