# The Proof of Innocence

#### alain

##### Older and Wiser
the paper has a vital flaw but the officer didn't spot it.

#### DMD

##### Active Member
the paper has a vital flaw but the officer didn't spot it.

#### alain

##### Older and Wiser
I think you are being sarcastic but internet forums are very bad at this. Anyway, look at Eq (8) on page 2. What is missing?

#### Dismantler

I think you are being sarcastic but internet forums are very bad at this. Anyway, look at Eq (8) on page 2. What is missing?
In the paragraph above is says the car comes to a complete stop then accelerates again...the velocity is zero or did I miss something?

#### Yike Lu

##### Finder of biased coins.
Anyway, look at Eq (8) on page 2. What is missing?
Air resistance!

In seriousness, the lack of initial velocity does not invalidate the analysis - if we assume that the car's deceleration is perfectly calculated to stop at S, then we can simply set t=0 to be the time at which it stopped. From there, the physics is symmetric about t=0 and so the graphs he produces are still valid since the processes of acceleration and deceleration are perfect mirror images of each other.

If I were to find a flaw, it's that human perception is sharper than he's stating. Half a second is very obvious if one is paying a ttention.

#### alain

##### Older and Wiser
Air resistance!

In seriousness, the lack of initial velocity does not invalidate the analysis - if we assume that the car's deceleration is perfectly calculated to stop at S, then we can simply set t=0 to be the time at which it stopped. From there, the physics is symmetric about t=0 and so the graphs he produces are still valid since the processes of acceleration and deceleration are perfect mirror images of each other.

If I were to find a flaw, it's that human perception is sharper than he's stating. Half a second is very obvious if one is paying a ttention.

I think it does. The car was already moving so Eq 8 becomes
$$x(t)=v_0t + \frac{1}{2}a_0t^2$$

#### Yike Lu

##### Finder of biased coins.
I think it does. The car was already moving so Eq 8 becomes

$$x(t)=v_0t + \frac{1}{2}a_0t^2$$
No, you missed the point of my response then. Yes strictly speaking you are correct, but you can instead simply view it this way:

Taking t=0 to be the point at which the car is completely stopped, the equation
$$x(t)=\frac{1}{2}a_0t^2$$
holds for t > 0.

For t < 0 the following equation holds
$$x(t)=-\frac{1}{2}a_0t^2$$

#### alain

##### Older and Wiser
I understand your point. However, using your point of reference, the car was moving on t < 0. The distances traveled are different if we considered t=0 the point where the car stopped, correct?

#### Yike Lu

##### Finder of biased coins.
Actually, the author wrote the whole paper assuming that the stopping point was t = 0, I just noticed that.

The distances traveled won't change for any choice of reference point for t, or of x, as t is arbitrary and x is as well, since the dynamical equations are the same. The only thing that will change is the specific values of x(t), v(t), however, we are only interested in what happens between times, not absolute time.

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