OK, well the way to do this is to fill your car with a rather low-pressure mixture of Deuterium (${}^2H$) and Tritium, (${}^3H$). The pressure needs to be low enough such that the mean free path of the atoms is about the length of the inside of the car, which I'll take to be 2 metres.

Then you put your foot on the accelerator really hard, and some of the Tritium atoms will fall from the front of the car and hit Deuterium atoms close to the rear windscreen. Apparently the collision energy needs to be about $0.1\,\mathrm{MeV}$: this is the kinetic energy that the Tritium atom needs to have gained (it's better for the Tritium to do the falling towards the Deuterium as it's heavier so it does not need to fall so fast).

The first thing to do is to check whether this needs the speed to be relativistic, as there's no way I'm doing this calculation if it does. So, using $E=mv^2/2$ or equivalently $v = \sqrt{2E/m}$, we want $v \ll c$ The atomic mass of Tritium is about $5\times 10^{-27}\,\mathrm{kg}$ and $0.1\,\mathrm{MeV} \approx 1.6\times 10^{-14}\,\mathrm{J}$. And plugging this in we get

$$v \approx \sqrt{\frac{3.2\times 10^{-14}\,\mathrm{J}}{5\times 10^{-27}\,\mathrm{kg}}} \approx 2.5 \times 10^6\,\mathrm{ms^{-1}} \ll c $$

So we're fine.

So now we can just use the constant-acceleration formulae we learn in school: $v = u + at$ and $s = ut + at^2/2$, and solve for $a$ in terms of $s$ and $v$, both of which we know. And the formula is

$$ a = \frac{v^2}{2s} $$

So $$a \approx \frac{\left(2.5 \times 10^6\,\mathrm{ms^{-1}}\right)^2}{4\,\mathrm{m}} \approx 1.6\times 10^{12}\,\mathrm{ms^{-2}}$$

Or, in other words $a \approx 1.6\times 10^{11}\,\mathrm{g}$. As well as breathing apparatus you will need a g-suit.

As you can tell, I am trying to avoid work today.

OK, well the way to do this is to fill your car with a rather low-pressure mixture of Deuterium (${}^2H$) and Tritium, (${}^3H$) (yes, this does not meet your requirements: the size of my envelope does not allow for calculations any harder than this one, sorry). The pressure needs to be low enough such that the mean free path of the atoms is about the length of the inside of the car, which I'll take to be 2 metres.

Then you put your foot on the accelerator really hard, and some of the Tritium atoms will fall from the front of the car and hit Deuterium atoms close to the rear windscreen. Apparently the collision energy needs to be about $0.1\,\mathrm{MeV}$: this is the kinetic energy that the Tritium atom needs to have gained (it's better for the Tritium to do the falling towards the Deuterium as it's heavier so it does not need to fall so fast).

The first thing to do is to check whether this needs the speed to be relativistic, as there's no way I'm doing this calculation if it does. So, using $E=mv^2/2$ or equivalently $v = \sqrt{2E/m}$, we want $v \ll c$ The atomic mass of Tritium is about $5\times 10^{-27}\,\mathrm{kg}$ and $0.1\,\mathrm{MeV} \approx 1.6\times 10^{-14}\,\mathrm{J}$. And plugging this in we get

$$v \approx \sqrt{\frac{3.2\times 10^{-14}\,\mathrm{J}}{5\times 10^{-27}\,\mathrm{kg}}} \approx 2.5 \times 10^6\,\mathrm{ms^{-1}} \ll c $$

So we're fine.

So now we can just use the constant-acceleration formulae we learn in school: $v = u + at$ and $s = ut + at^2/2$, and solve for $a$ in terms of $s$ and $v$, both of which we know. And the formula is

$$ a = \frac{v^2}{2s} $$

So $$a \approx \frac{\left(2.5 \times 10^6\,\mathrm{ms^{-1}}\right)^2}{4\,\mathrm{m}} \approx 1.6\times 10^{12}\,\mathrm{ms^{-2}}$$

Or, in other words $a \approx 1.6\times 10^{11}\,\mathrm{g}$. As well as breathing apparatus you will need a g-suit.

As you can tell, I am trying to avoid work today.