# Total distance

1. Jun 30, 2004

### vikasj007

A car, initially still, accelerates up to a maximum speed of v, and then slows down to a stop again. The plot of the velocity over the time (t seconds from start to stop) is a semicircle.

(i'm sorry i cannot post the diagram, but try to imagine it, it is a simple semicircle where the topmost point is the max. veloicity and t is the total time elapsed.)

Now as all mathematicians know, it is often useful to find the area under the curve, in this case to find how far the car has travelled. One mathematician claims that since the radius of the semicircle is v, the distance travelled is (pi/2)*v^2. Another disagrees, saying that t is the diameter of the semicircle, so the distance is actually (pi/2)*(t/2)^2. Which one is right?

if you know the answer, do not spoil it for others, to check if u r correct or not, send me a PM.

2. Jul 4, 2004

### Ursole

I thought this was a forum.

3. Jul 5, 2004

### AKG

Don't read if you don't want it spoiled:

Let V and T be the max speed and total time, respectively. We can model the plot of velocity vs. time as the ellipse:

$$\frac{v^2}{V^2} + {(t - T/2)^2}{(T/2)^2} = 1$$

Of course, the graph will only be the top half. Now, the area of half of this ellipse will be the distance, D:

$$D = 1/2 \times V \times T/2 \times \pi = \frac{VT\pi}{4}$$

4. Jul 5, 2004

### vikasj007

that's right AKG.

5. Jul 8, 2004

### gerben

I do not get this. Do you just mean that the plot is not half a circle?

Which mathematician is right? (that was the question)
I mean it is obvious that both should be right because v = t/2, that just has to be the case if the plot is half a circle (which would of course be physically impossible, but well that does not matter if it gives an interesting brain teaser).

also AKG's answer seems to imply that both mathematicians are right, because:
(v*t*pi)/4 = (pi/2)*(t/2)*v = (pi/2)*(t/2)^2 = (pi/2)*v^2

so what's the catch???? (how does AKG's answer solve this "brain teaser"?)

Last edited: Jul 8, 2004
6. Jul 8, 2004

### AKG

Think of a velocity-time graph, drawn on grid paper. The velocity axis goes up in increments of 5 m/s [forward], and the time axis goes up by increments of 1 s. Is there something wrong with this graph? No. You could also equally make the time axis go up by 5s. Now, if this were the case, your graph originally would horizontally shrink by a factor of 5 in appearance. Basically, the "catch" is that a velocity time graph can look like a circle, or an ellipse, depending on the scale. And since there is no one right scale, it's fair to call it a circle. For one thing, "time" doesn't have to be in seconds, it can be in 2.3seconds. So, let's say that the velocity after 2.3 seconds is 1 m/s. Now, let the unit of the velocity axis be m/s, and let the unit of the time axis be 2.3s. This way, you'd get a point at (1,1). Since the units of seconds are arbitrary, you can take a velocity-time graph and fool around with the units so that it actually has the equation of a semi-circle. Since you know the units may not simply be 1s, (it could be 2.3s), you have to, instead, treat the graph as a general ellipse, because in units of 1m/s and 1s, it may be an ellipse. Then you can solve for the area.

EDIT: So both mathematicians are right only if the units for t are 1s, and the units for v are 1 m/s [forward].

Last edited: Jul 8, 2004
7. Jul 8, 2004

### arildno

Personally, I think neither of them are correct since their units are wrong (unless their using non-dimensional units of course..)

8. Jul 8, 2004

### NateTG

No, it's common practice to use a particular convenient scale to plot something. The trick is that the dimensions might be a bit odd. Maybe time is measured in years, and velocity is measured in furlongs per day.

So, I will use my own custom units of time and distance:
I'll use zeconds($$z$$) for time and neters ($$n$$) for distance. Let's say that the total time is $$t$$ seconds, and the peak velocity is $$v$$ meters per second.

I'll define zeconds so that the total time is two zeconds i.e. $$1z=\frac{t}{2} s$$ seconds, and define neters so that the peak velocity is one neter per zecond i.e. $$1n = \frac{v \times t}{2}$$.

Now, it's easy to see that the distance will be $$\frac{\pi}{2}$$ neters, or equivalently, $$\frac{v t \pi}{2}$$ by applying the conversion factors. Which is, not accidentaly, half the area of an elipse with axis lengths of $$v$$ and $$t$$.

Using an appropriate scale, any eliptical plot can be converted to a circular one.

I would go with both, or neither, depending. The answers are, after all, guaranteed to be numerically equivalent.

Last edited: Jul 8, 2004
9. Jul 8, 2004

### AKG

That's essentially what I said. My point was that t = 1 could mean 1 second, or 2.3 seconds (or 1 year), depending on if the unit for times measuerment was the second, the year, or the 2.3-seconds.

10. Jul 8, 2004

### NateTG

Perhaps you meant one thing, and typed another. The former quote is explicitly about units being meters and seconds, and not about quantity.

11. Jul 8, 2004

### AKG

Yeah, the former quote was mistaken on my part.

12. Jul 8, 2004

### gerben

I agree that you can change the shape of the v,t plot by choosing different units, but the question was simply: what is the traveled distance given that the v,t plot is half a circle?
With regard to the units this means that v and t are given in such units that the plot is half a circle. Therefore v=t/2, and so there is no "teasing the brain" at all, I think. You just have to multiply the square of the number (v=t/2=radius) with pi/2.

(the units of the distance that is thus calculated depends of course on the units that were used to draw the plot as a semicircle, but this has nothing to do with the question)

EDIT: I implicitly assumed that the mathematicians are physically minded enough to choose units for which the following applies:
units_used_for_Distance <=> units_used_for_Time * units_used_for_Velocity

Last edited: Jul 8, 2004
13. Jul 8, 2004

### AKG

First of all, v=t/2 is meaningless because the units don't match up. Both mathematicians' approaches yield technically meaningless results. Now, is the result at least numerically correct? In other words, if we get rid of units, are we okay? No. Let's say we go with v being the radius. It could be 3.6 kilometer/hour or 1 meter/second. If we ignore units, we get two different numbers, and our final answer could be either one of these numerically, which is unacceptable. Since we do need to take units/dimensions into account, we can't even expect to get a numerically correct answer with that approach.

14. Jul 8, 2004

### gerben

Oh you responded when I just edited post #12, I think the addition I made:
" EDIT: I implicitly assumed that the mathematicians are physically minded enough to choose units for which the following applies:
units_used_for_Distance <=> units_used_for_Time * units_used_for_Velocity"
makes my point clear, both mathematicians are right

15. Jul 8, 2004

### AKG

gerben

I think you may be right. My solution and the solution of those mathematicians are the same if they decide the units in the way you do (of course, that's a rather abnormal way to do it, choosing the units after the calculation, the dimensions should work out in the calculations themselves) as long as v and t are numerically equivalent, which I suppose is implied in the fact that the graph is a circle (and we assume that the units on the graph correspond to the units of measurement for t and v).

16. Jul 8, 2004

### gerben

AKG

I do not think it matters how they choose units, they will both get the same answer for the distance (because t/2 is equal to v if the plot is a semicircle) even if de velocity was in 3.6 kilometer/hour and the time in seconds. In that case the distance that they (both) will get is in units that are 3.6*60*60 times as long as a kilometer, which would also be the case for your formula. If more usual units were used like m/s for the velocity and s for the time (and the plot is a semicircle) the distance will also be in more usual units (of length 1 m).

However nothing is said about units, so although they may be wrong if they interpret one of the numbers (v, t or d) in the "wrong" units that has nothing to do with the question. The question is not about units, its just .... well I do not get how it could be a "brain teaser"

A similar "(non) brain teaser" would be:
Two 'mathematicians' were looking at a large wall that was made out of large cubic bricks. The wall is 4 bricks wide and 4 bricks high, one of them said the frontal area of that wall is equal to width^2, the other argued that it was equal to height^2. Who is right?
(in this case you will also get wrong answers is you do not express the width and the height in corresponding units and interpret the result as if they are in the same units)

AKG, your suggestion would be: “Its is equal to width*height”

Last edited: Jul 8, 2004
17. Jul 9, 2004

### vikasj007

So, i believe that this question is generating a lot more debate than it was supposed to. well, this was just a simple question to catch people off gaurd(and also those who don't know about units and dimensions). i think the cause for the confusion was my inability to post the graph which would have made the situation clear. but anyways, here if we consider the answers given by both the mathematicians, dimensionally they are both incorrect, as the unit for their answer should be of distance, which was not the case with any of the two.

i hope that clears up the argument.