The speed of tachyons The other day I was wondering: When a tachyon is coming towards you faster than the speed of light, will you see it before it hits you? Then I thought of course not, since the light waves aren't traveling faster than the tachyon then how could you see it before it hits you? Now I thought today, if an tachyon is traveling away from you faster than the speed of light, would you see it?
If you fire a ball at an initial velocity of 20mph south out of a car that is going 50mph north, the final velocity of the ball would be 30mph north, is this also how light acts when the initial velocity of the object it is reflecting off is not equal to 0? 
So in my case, if the speed of light were 100mph (dummy math) and a tachyon was traveling at 110mph north that means the light reflecting off the tachyon would be traveling at 10mph north, so then really would you be able to see it?
More generally, how does relativistic addition of velocities work for tachyons?
update:
This question is a hypothetical question: IF tachyons exist, then what would happen? After a few hours of research I see why a usual massive object CAN'T travel faster than (or even reach) the speed of light, but this question is about tachyons.
 A: If a tachyon starts from where you are and goes away at faster than the speed of light, you will see the photons it emits earlier than it actually departs.
So you will see all these photons coming as if the tachyon were coming toward you at a speed slower than light, and then bang, the tachyon leaves.
In fact, the faster it is going away, the slower it appears to be arriving.
EDIT: You can just tell this from a space-time diagram:
T  s   C
| /   /
|/   /    f
| s /    /
|/ /  f /
| /  / /
|/__/_/___X
|  / /
| / /
|/ /
| /
|/

Here, the time T axis is vertical and the space X axis is horizontal.
Line C represents the speed of light. Photons move parallel to that line.
If something is moving away from you slower than light, it is a diagonal line falling in the slow (s) region. When it emits photons, they travel parallel to C, so each one arrives back at you at a later time.
That's the normal behavior that you're used to.
If something is moving away from you faster than light, it is a diagonal line in the fast (f) region. When it emits photons, they travel parallel to C, and thus arrive back to you at a negative time, relative to when the object left you.
In fact the faster it's moving (closer to horizontal) the earlier its photons will arrive (negative T).
The slower it's moving (closer to C) the more its photons will appear to come all at once, just before it "departs".
A: As the commenter already indicated, there is no way for masses to reach a speed faster than the speed of light, $c$ (or even as fast as $c$). 
The analogy with the cars does not hold when you go to velocities close to $c$. Those velocities are called relativistic velocities. The Galilean transformations for low velocities do not hold any longer, instead, you should use the Lorentz transformations. This based on Einsteins special relativity theory.
For example, suppose you have an object at a distance of $1$ lightyear, moving with velocity $0.99c$ away from you.
With classical physics you would calculate that the speed of the photon is $0.01c$ towards you, so it would take the light $100$ years to reach you.
One of the postulates of Einstein, is that the speed of light is always $c$, irrespective of the frame of reference. So, the light photon emitted will travel with $c$ towards you, so you would be able to see the object after one year (instead of $100$).
A: If you fire a beam of photons at an object receding away from you at a speed greater than the speed of light, your photons will never reach it to reflect off it
OTOH, if such an object emits photons, you should eventually be able to see the object as it was at the time the photon was emitted.
A: The short answer is that no, light doesn't behave like a ball being thrown out of a moving car.  The speed of light in a vacuum, $c$, is a universal constant, meaning that all light always travels at the same speed, independent of the speed of the thing that emits it. So you would be able to see your faster-than-light object moving away from you, because its light would be travelling towards you at $c$.
The slightly longer answer involves explaining that the speed of light is also independent of the speed of the person observing it.  This seems impossible at first, but Einstein realised it was possible if you re-think the nature of space and time, and that's the basis of special relativity. Special relativity ultimately shows you that (i) it's impossible to accelerate an object any faster once it gets to the speed of light, and (ii) if objects could be created travelling faster than $c$ (thus avoiding the need to accelerate them), it would be possible to use them to send messages back in time, causing paradoxes - so it's very unlikely that this is possible.
A: The issue of tachyonness is a red herring; and the question is entirely a matter of simple computation, that you should do for yourself.
Here it is.
Let $x = v t$ describe the motion of an object in a direction parallel to the $x$ axis, where $t$ is the time, where the origin on the $x$ axis is where you are at and where time $t = 0$ is when the object is where you are at. It is moving with a speed $v$, which is negative if it is toward you, positive if it is moving away from you.
We'll also assume you're staring in the direction of increasing $x$. A signal moving at the light speed $c$ has a motion given by $x = b - c t$, if it is coming at you from the direction of increasing $x$. If it is at the location of the object at $(x, t) = \left(x_0, t_0\right)$ and at your location at $(x, t) = \left(x_1, t_1\right)$, then
$$
x_0 = b - c t_0, \hspace 1em x_0 = v t_0,\\
x_1 = b - c t_1, \hspace 1em x_1 = 0.
$$
Therefore $(v + c) t_0 = b = c t_1$, and:
$$(v + c) x_0 = (v + c) v t_0 = v (v + c) t_0 = v c t_1.$$
The motion you see is given by the stream of signals and is therefore described by $(x,t) = \left(x_0, t_1\right)$. You could do this extra carefully by trying to separately account for left and right eyes and use triangulation to determine the "apparent" distance in place of $x_0$, but I don't think that extra complication will significantly affect the issue (maybe except when it's close), so we'll just use $x_0$, itself.
Thus, assuming $v ≠ -c$, and $v ≠ 0$, we have
$$x_0 = \frac{v c}{v + c} t_1 = \frac{c}{1 + c/v} t_1 = \frac{v}{1 + v/c} t_1,$$
and you see the object moving with a speed $c/(1 + c/v) = v' = v/(1 + v/c)$.
If $v = 0$, then $x_0 = v t_0 = 0$ and you see the object up right next to you, so $v' = 0$.
If $v = -c$, then $b = 0$ and, thus, $t_1 = 0$, and that's the only signal you'll ever see, and you'll only see it in a flash at time 0, and $v' = ∞$.
If $v > 0$, then $0 < v' < c$. If $-c < v < 0$ then $v' < 0$. If $v < -c$ then $v' > 0$. If $v = ∞$ then $v' = c$. An object going at the speed $v = ∞$ will appear to you to be moving away from you at light speed - even if you turn around and look at it in the other direction.
By the way, notice that I actually said nothing about Relativity in the determination of $v'$. Relativity is also a red herring to the question; it's the same answer as in non-relativistic physics. You get the same expression for $v'$, regardless of which paradigm you are in, it is paradigm-independent, and the question actually has nothing to do with Relativity, per se, but merely with signal processing. You could just as well take any other speed $V$ in place of $c$ (e.g. the speed of light in water) to determine the visual effect of an object's motion, in which case you'll get $v/(1 + v/V) = v' = V/(1 + V/v)$ in place the expression previously derived - again, independent of paradigm. So, even light speed is a red herring here.
