has 1500W, which means it draws 31A at 48V at full load. The inrush current can be higher, that's your 40-50A. However, this doesn't mean the motor needs 40-50A to start and 31A to run, it's just the absolute maximum. At lower voltage, it will sink a lower current and have lower power.
Remember, a combustion has its maximum power near the red line, but it's rarely driven at that RPM, even not when accelerating.
The solar panel
has an maximum output of 300W. But only under best conditions, which means no clouds, no fog, the sun as high as possible in the sky, and the panel pointing exactly to the sun.
In a few days, we have September 21th, where the sun is exactly above the equator, and the latitude of your location gives the angle between the sun and a vertical line at noon. New york for example is at 40°N, so the angle will be 40° (or: The sun will be 90°-40°=50° above the horizon). If you place the panel on the ground, it will only give cos(40°)*300W=0.77*300W=230W. And this doesn't take into account that the light had a longer way through the atmosphere and has been attenuated more than at the equator. (And it's high noon. Not morning, not afternoon)
Further more, a solar panel is a constant current source. With a load of 7.7Ohm, it will deliver 48V and 6.25A (=300W). With a higher load, the current will still be 6.25V, but the voltage will break down, which means the power delivered will break down, too.
If you apply a smaller load, the voltage will rise, but the current will fall. Again, the power decreases.
So, this 48V, 6.25A are the maximum power the panel gives with the right load connected. And the 48V are about 80% of the voltage when no load is connected, which is about 60V.
First, if you put two caps in series, the voltage rating of both can be added, but the capacity is
Ctot = 1/ (1/C1 + 1/C2)
which means the combination of two 58F caps gives a 29F cap. If you connect three caps in series as Paulster2 suggested, you get just 19.3F. But this doesn't really matter in terms of stored energy. Each cap can store
and a combination of three caps in series will provide just 1/3 of capacity, but 3x the max voltage. It turns out three caps will store three times the energy a single cap can store, as expected.
The max energy of a single cap is 7.4kJ (for comparison: a 40Ah car battery has 1728kJ) and it will take roughly 7400J/300W=25s to charge it at the max. power of the solar panel. More caps need more time to charge.
Now, let's say the motor pulls 1500W, whereof 300W are supplied by the panel. This makes 1200W for the caps, and a single cap will last for 7400J/1200W=6s.
A few warnings:
Your caps (this ones?) are rated for 16V, and absolute maximum voltage is 17V, which isn't much more. You should never exceed that voltage, otherwise the cap could be damaged or even burn/explode.
If you connect three caps in series to get 48V and directly connect them to the panel without protection, they will be charged to the open circuit voltage of 60V and BAAANG!
Next, large caps have a lousy precision in capacity. Your caps have a capacity of 58-70F. If you connect three in series to 48V, two having 70F and one 58, the big ones will charge to 15V and the small one to 18V - and BAAANG!
So I would recommend to either not put the caps in series - though this results in high currents - or not charge them to the theoretically rated voltage.
So it's clear you shouldn't connect the panel to three caps in series, and it's also clear that you shouldn't connect them to the motor directly - their voltage will decrease and they will not provide max. power to the motor.
You will definitely need some DC-DC converters, and my calculations assume this, too. Converters always have some inefficiency, which heavily depends on input/output voltages and currents. High currents are always a little problematic, and when the caps are almost empty, they must provide a giant current to provide the requested power. I'm not able to give any numbers, but in principle you have to take into account that whenever energy is transferred from the panel to the caps and from the caps to the motor, a fraction is wasted by the converter.
(And finally, you want to control the power of the motor, too)
Though your motor has a max. power of 1500W, you do not need to provide 31A at 48V unless you really want the entire 1500W. Typically, you will need much less power.
The max power of the panel is 300W which already is a bit low, but you should expect to get waaaay less out of it.
This caps can for sure help to buffer energy and boost the motor when more power is demanded or some clouds show up , but they must be charged, first. I guess they could count as battery in your ... contest unless they are not charged for the start.
Your comment is a little confusing, I'm still not sure what exactly you are planning to do. You also mention a battery, though the question says you have to remove the battery...
Anyway: You will need a DC-DC converter to transfer energy from the caps to the motor. On the other side, you need to charge the caps with the excess of power from the panel. And when you need max. power and the caps are empty, you want the entire power from the panel flow into the motor, and not into the caps.
Further more, as I wrote, when you try to get more power from a panel than it can deliver, it actually delivers less power.
Unfortunately, I don't know any electronics you can directly use, though it for sure exists.