Required cable thickness (and thus weight) is proportional to current, not voltage. Conductors to carry megawatts of electricity could be fairly light as long as the voltage is extremely high and the amperage relatively low.
The issue with acceleration using wheels is gravity. You are limited to 1g, the downforce on the wheels, which you use for traction to push the vehicle forward. Spoilers increase the downforce on the vehicle beyond 1g, but the energy required to do that is proportional to the drag on the spoiler.
Short of a rocket engine, the way to accelerate beyond 1g with a wheeled vehicle is to have a downforce fan. That can be powered by combustion or by electricity. I don't know which would offer the highest power to weight ratio.
I also feel like 800V cars also maintain their maximum power better on lower SoC than 400V systems.
[1] https://www.copper.org/publications/pub_list/pdf/A6192_Elect...
This electric vehicle is accelerating to 100kph (~28 m/s) in less than 1.5 seconds, which is an average acceleration of about 2g, if I am not mistaken.
Looking around, I have seen 4.0 given as a representative friction coefficient for drag racing tires, which would imply a corresponding peak acceleration of 4g without aerodynamic assistance (again, if I am not mistaken.)
The acceleration you can reach is mostly limited by the peak current and the force constant, but the maximum speed you can reach at that acceleration is mostly limited by the voltage you have available because of the back-EMF. A simplified formula is that the voltage you need is R * I + v * BEMF. The higher the voltage is, the better the insulation between the windings needs to be. You need to find a balance between the different properties.
Let's say you can control voltage, cross section area A.
Now, you want a constant power across a load P = V * I.
Your net resistance R is C + D * L / A where C and D are constants and L is your cable length.
I is proportional go V * A / (C * A + E) where E is also a constant.
So your load power is proportional V^2* C * A / (C * A + E) which is proportional to V^2 * A / (A + F) where F is also a constant.
With a large "enough" A, this is effectively V^2. With a small A, this is V^2 * g where g is A / F. So the smaller the area you have the more power you are wasting (roughly equal to V^2 * (1-g) which is heat in the wires).
So the smaller area you have, the less efficient your power delivery. And juicing up your source power is a lot more expensive than juicing up your source voltage.