How fast would a car be propelled by this capacitor ?

Lets keep this simple.  

I have a 3000 lb car
I have one of these:

Maxwell BMOD0063 P125 B33 22 63 Farad 125 Volt Ultracapacitor Module

Assuming a perfect state, How many of these would it take to propel the car to 50 mph ?
Assuming a perfect state, How fast would this propel the car assuming full discharge ?

Can you show me that math ? 30 years ago I remeber we had to convert lbs > KG ?  

I asked almost the same question before however I would like to understand if dollar for dollar the 125 volt ultracap is going to be cheaper than the 48 ultracap ?

the 48 volt ultracap was 83F and  1000$.  This cap is 125 volt, 63 F and 3000 $



TIMFOX123Asked:
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ozoConnect With a Mentor Commented:
3000 lb = 1360.777 kg
50 mph = 22.352 m/s
1360.777 kg * (22.352 m/s)^2 / 2 = 339930.22 J
63 F * (125 volt)^2 / 2 = 492187 J
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TIMFOX123Author Commented:
Assuming:

63 F * (125 volt)^2 / 2 = 492187 J
then
89 F * (48 volt)^2 / 2 =   102528 J

I must be doing my math wrong or something.  
4.800513031
the 3k dollar ultracapacitor is 5 times as powerfull as the 1k dollar capacitor.   They are just the same inside but the number of capacitors is more in the higher voltage one.



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TIMFOX123Author Commented:
the 48 volt ultracap was 83F and  1000$.  This cap is 125 volt, 63 F and 3000 $
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aburrConnect With a Mentor Commented:
energy = (1/2) * C * (V)^2
for 48 V cap  95616
for 125 V cap 492167
It would take 5 of the 48 V caps to contain slightly more energy  than  one 125 V cap. So the 125 cap is the better buy.
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aburrConnect With a Mentor Commented:
ozo showed that one 125 v cap would do nicely. four 48 v caps would also work
Again the 125 v is cheaper

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TIMFOX123Author Commented:
thanks !!
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