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There are two ways to interpret this question, since in relativity theory, acceleration is relative, too.If an object accelerates at [math]100~{\rm m}/{\rm s}^2[/math] relative to a stationary observer, it would, of course, reach the speed of light in exactly [math]2997924.58~{\rm s}[/math]. However, near the end of this time interval the force required to maintain that constant acceleration increases beyond limit to infinity, which is of course quite impossible.If an object accelerates at [math]100~{\rm m}/{\rm s}^2[/math] as measured in its own (accelerating) reference frame, then it will take forever. The force applied would be constant, but to a stationary observer, the actual acceleration of the object will appear to diminish as the object approaches (but never reaches) the speed of light.

Indeed, you can get rid of the unit of time by replacing it with a unit of length. So for instance, instead of 5 seconds, you could write 1,498,962,290 meters. Note that in these units, the speed of light will be 1.But wait…you can do more. You can also get rid of the unit of mass. One way to do it is by using Newton’s constant of gravitation, [math]G = 6.674\times 10^{-11}~{\rm m}^3/{\rm kg}/{\rm s}^2[/math]. Replacing seconds with meters gives us [math]7.426\times 10^{-28}~{\rm m}/{\rm kg}[/math]. That is to say, you can get rid of the kilogram, too, by writing [math]7.426\times 10^{-28}~{\rm m}[/math] in its place. In these units, by the way, Newton’s constant of gravitation will also be 1.But wait… there is something else you can do! Let’s look at Planck’s constant, or rather, the so-called reduced Planck constant: [math]\hbar=1.055\times 10^{-34}~{\rm m}^2{\rm kg}/{\rm s}[/math]. We already know how to replace kilograms and seconds with meters, which gives [math]\hbar = 2.613\times 10^{-70}~{\rm m}^2[/math]. So what if, instead of meters, we used [math]\sqrt{\hbar G/c^3}=1.617\times 10^{-35}~{\rm m}[/math] as our unit of length?Well, in this case, all three constants have the value of unity: [math]c=G=\hbar=1[/math]. And we got rid of all units.Theoretical physicists like to do calculations this way. They call this formalism “dimensionless”, and the units “natural units”.So theoretical physicists might tell you, for instance, that the Schwarzschild radius of a star is [math]r_s=2M[/math]. But that cannot be right… a length cannot equal a mass! So maybe it is [math]r_s=2GM[/math]? A bit closer (at least we got rid of the mass) but now we have a length of the left-hand-side and length cubed divided by time squared on the right. So what if we write [math]r_s=2GM/c^2[/math]? Hah! That’s length equals length, so it looks right! This is what these theorists mean when they say that we are “restoring units”.Once you get used to it, it greatly simplifies the writing of equations in relativity theory, quantum field theory and related fields, and the process of “restoring units” is (almost) always unambiguous and can be done at the very end.