I read somewhere that $E=mc^2$ shows that if something was to travel faster than the speed of light then they would have infinite mass and would have used infinite energy.
Nope, not true. For a couple of reasons, but first, let me explain what $E = mc^2$ means in modern-day physics.
The equation $E = mc^2$ itself only applies to an object that is at rest, i.e. not moving. For objects that are moving, there is a more general form of the equation,
$$E^2 - p^2 c^2 = m^2 c^4$$
($p$ is momentum), but with a little algebra you can convert this into
$$E = \gamma mc^2$$
where $\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$. This factor $\gamma$, sometimes called the relativistic dilation factor, is a number that depends on speed. It starts out at $\gamma = 1$ when $v = 0$, and it increases with increasing speed. As the speed $v$ gets closer and closer to $c$, $\gamma$ approaches infinity. Armed with this knowledge, some people look at the formula $E = \gamma mc^2$ and say that, clearly, if a massive object were to reach the speed of light, then $\gamma$ would be infinite and so the object's energy would be infinite. But that's not really true; the correct interpretation is that it's impossible for a massive object to travel at the speed of light. (There are other, more mathematically complicated but more convincing, ways to show this.)
To top it off, there is an outdated concept called "relativistic mass" that gets involved in this. In the early days of relativity, people would write Einstein's famous formula as $E = m_0 c^2$ for an object at rest, and $E = m_\text{rel}c^2$ for an object in motion, where $m_\text{rel} = \gamma m_0$. (The $m$ I wrote in the previous paragraphs corresponds to $m_0$ in this paragraph.) This quantity $m_\text{rel}$ was the relativistic mass, a property which increases as an object speeds up. So if you thought that an object would have infinite energy if it moved at the speed of light, then you would also think that its relativistic mass would become infinite if it moved at the speed of light.
Often people would get lazy and neglect to write the subscript "rel", which caused a lot of people to mix up the two different kinds of mass. So from that, you'd get statements like "an object moving at light speed has infinite mass" (without clarifying that the relativistic mass was the one they meant). After a while, physicists realized that the relativistic mass was really just another name for energy, since they're always proportional ($E = m_\text{rel}c^2$), so we did away with the idea of relativistic mass entirely. These days, "mass" or $m$ just means rest mass, and so $E = mc^2$ applies only to objects at rest. You have to use one of the more general formulas if you want to deal with a moving object.
Now, with that out of the way: unfortunately, the passage you've quoted from Hawking's book uses the old convention, where "mass" refers to relativistic mass. The "equivalence of energy and mass" he mentions is an equivalence of energy and relativistic mass, expressed by the equation $E = m_\text{rel}c^2$. Under this set of definitions, it is true that an object approaching the speed of light would have its (relativistic) mass approach infinity (i.e. increase without bound). Technically, it's not wrong, because Hawking is using the concept correctly, but it's out of line with the way we do things in physics these days.
With modern usage, however, I might rephrase that paragraph as follows:
Because energy contributes to an object's inertia (resistance to acceleration), adding a fixed amount of energy has less of an effect as the object moves faster. This effect is only significant to objects moving at speeds close to the speed of light. At 10 per cent of the speed of light, it takes only 0.5 per cent more energy than normal to achieve a given change in velocity, but at 90 per cent of the speed of light it would take twice as much energy to produce the same change. As an object approaches the speed of light, its inertia rises ever more quickly, so it takes more and more energy to speed it up by smaller and smaller amounts. It cannot therefore reach the speed of light because it would take an infinite amount of energy to get there.
Disclaimer: all I've said here applies to a fundamental particle or object moving in a straight line. When you start to consider particles with components which may be moving relative to each other, the idea of relativistic mass kind of makes a comeback... kind of. But that's another story.
For any general body, the energy (in a relativistic framework) is given by,
$E=\sqrt{m_0^2c^4+p^2c^2}=m(v)c^2$
For a photon, the rest mass $m_0=0$, and it's energy $E=pc$. The relativistic mass $m(v)$ is not zero for the photon. Also, the momentum $p=m(v)v$, not $m_0v$, so the formulas are consistent too.
From the above expression for energy, we may arrive at the formula your friend gave you, which is
$m(v)=\dfrac{m_0}{\sqrt{1-\dfrac{v^2}{c^2}}}$
But in the case of a photon, as $m_0=0$, then you don't arrive at the above formula. Putting rest mass as zero, and $v=c$ in the expression of energy, we get
$m(v)=E/c^2=p/c=m(v)v/c=m(v)$, as $v=c$ for a photon.
Therefore, no division by zero. In the case of light, there is no rest mass, and we speak about energy and momentum of light, in place of mass. Because in this case, they are directly related to each other (proportional).
Edit: To those who downvoted it, please inform me of my mistakes so that I can edit my knowledge for the better. Thank You!
Best Answer
Your second equation is incorrect. In both Newtonian Mechanics and Relativistic Mechanics, force is the time rate of change of momentum:
$\vec F = \dfrac{d \vec p}{dt}$
Where, in special relativity, momentum is:
$\vec p = m \dfrac{\vec v}{\sqrt{1 - \frac{v^2}{c^2}}} = m \gamma \vec v$
and $m$ is the invariant mass.
Thus, when the time derivative is taken, by the product rule:
$\vec F = m(\gamma \vec a + \dot \gamma \vec v)$
Note that, in general, the force vector is not parallel with the acceleration vector!
If the force vector is always parallel with the velocity vector, the force equation simplifies to:
$F = m \gamma^3 a = m \dfrac{a}{(1 - \frac{v^2}{c^2})^{\frac{3}{2}}}$
Now, when you write:
you must actually be more specific. It appears that you're thinking about coordinate acceleration however, there is also the acceleration as measured by accelerometers (the proper acceleration) and this distinction is often not appreciated by SR "newbies".
While it is possible for the proper acceleration to be uniform, it isn't possible for the coordinate acceleration to be uniform as that would require unlimited force.