# How fast do I need to go in order to avoid being seen by the Police?

I was driving down the road at roughly the speed of traffic. I saw a police officer parked on the side of the road, and also noticed that a Semi was traveling in the lane right next to him. This got me thinking, is it possible to avoid being seen from the officer by using the Semi Truck and how fast would I have to go?

Let's say the officer is parked on the side of the road and the semi is traveling on the lane next to the officer:

Let's also assume that the truck is traveling at roughly constant 65 mph. Also assume that I see the officer and line up my car to block the officer's view via the semi at roughly a mile away. (Ya I know a little far).

What I want to know:

• How fast do I have to go to avoid being seen by the officer?

• Is this speed constant or variable?

Note: I am in NO way advocating speeding on the highways and breaking the law. This was just was just sparked as it somewhat happened while I was driving down the road.

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How will you know the cop's there in the first place? – yatima2975 May 24 '11 at 21:35
@yatima2975 Although this is highly hypothetical radar detection might be a means of detecting him. – KronoS May 24 '11 at 22:30
Considering that the speed limit for trucks is usually 10 mph lower than the speed limit for passenger vehicles on most highways in the US, this probably isn't a good plan if you were to want to speed :P – nhinkle May 24 '11 at 22:34
@nhinkle on the contrary by answer of @Spencer... you just have to speed really fast... – KronoS May 24 '11 at 22:56

What you want to do is keep the angle between your direction of motion and the line of sight to police car the same as the angle between the truck's direction of motion and the truck's line of sight to the police car.

In other words, we want to keep $a1=a2$ in the picture above.

This is a problem in similar triangles. The answer will be that the ratio of your motion to your distance to the cop will have to be the same as the ratio of the truck's motion to the truck's distance to the cop. This can be seen from the following:

We note that $$\cot{(a1)}=\frac{\textrm{Truck's Speed}}{d1}$$ $$\cot{(a2)}=\frac{\textrm{Car's Speed}}{d2}$$

Set $a1=a2$ so we are always hidden behind the truck and solve for $\textrm{Car's Speed}$: $$a1=a2$$ $$\cot{(a1)}=\cot{(a2)}$$ $$\frac{\textrm{Truck's Speed}}{d1}=\frac{\textrm{Car's Speed}}{d2}$$ $$\textrm{Car's Speed}=\frac{\textrm{Truck's Speed}\times d2}{d1}$$

So, suppose the lanes are the same size (10 feet wide, say), and the cop is 5 feet off of the highway. Then, $d1=5\textrm{ feet}$, $d2=15\textrm{ feet}$. The speed you need if the truck is going 65 miles per hour is $$\textrm{Car's Speed}=\frac{65 \textrm{mph}\times 15}{5}=195\textrm{ mph}$$

Edit: Some concerns were raised in the comments that this treats the truck as a point. This turns out not to matter. Here's a second picture like the first, but now we have a zone (colored in green) which the truck covers.

The green triangle gives you a little bit of wiggle room, since you can be covered by the front of the truck or the back or anything in between. However, the total size of your wiggle room does not change while you move (in other words, it doesn't depend on a4). As a result, it should be pretty clear that this doesn't change things much at all - we can think of it as two point-size trucks going at the same speed, and we have to stay between them. Of course, this will give exactly the same answer as the first case - it's really just like hiding behind one point-sized truck.

There is actually one small change, as David notes: If you start out covered by the front of the truck, you can go a little slower than the 195 mph cited above, because you can slowly slide back until covered by the back of the truck. However, if the length of the truck is $L_{tr}=40\textrm{ feet}$ (say), then this change in the velocity is quite small.

For example, suppose that we slide back 40 feet from the front of the truck to the back over the course of a mile. We're going 195 miles per hour, so it takes us 18.5 seconds to go one mile. In those 18.5 seconds, we move 40 feet relative to the truck; this is a speed of about 1.5 mph. So, we can go 1.5 mph slower if we start at the front and go to the back over a mile; taking this into consideration, we get that the speed needed is actually 193.5 mph.

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Very good explanation! +1 – KronoS May 24 '11 at 19:45
Nicely done :-) I would point out that in practice, you could be going a little slower if you start out hiding behind the front of the truck and end up hiding behind the back, but then you would need to factor in how long you want to remain hidden from the police car. – David Z May 24 '11 at 20:13
There's also the issue that this is developed for a single point on the truck... I could in theory be somewhat further up or behind the truck, just as long as the truck covers. – KronoS May 24 '11 at 22:57
@Kronos, @David I've addressed your comments with an edit. – spencer nelson May 25 '11 at 20:43

I would rather the cop be drawn on the left. Cop is the origin. Then both lanes intersect with the x-axis, at point $a$ for the truck's lane and $b$ for your lane and $a<b$. Now, some line of $y=m x$ intersects all 3 participants. The truck is moving with motion that can be written as

$$y_{truck}(t) = m(t) a$$

. We know

$$\frac{d y_{truck}}{dt}=65mph$$

, so we know (differentiate both sides)

$$\frac{d m}{dt} = \frac{65 mph}{a}$$

, and it's easy to then see that

$$\frac{d y_{you}}{dt} = 65 mph \frac{b}{a}$$

done. Yes it is constant and is larger by the most obvious fraction you could think of.

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It reduces down to triangle similarity.!

illustration of triangles

$X$ and $x$ are functions of time and $a$ and $b$ aren't. So we have:

$\frac{X}{x}=\frac{a}{b}$

$X=\frac{a}{b}x$

Deriving in relation to time:

$v=\frac{b}{a}V$

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