# The Physics of the Batmobile (Frictionless!!!)

As promised, I looked into how fast the Batmobile would go on a frictionless plane instead of the roads of Gotham. To compare the difference, I used everyone's loving friend, Energy.

u_{k}= .8

a_{B}= 60 mph in 2.4 seconds (after some conversions....)= 11.17 m/s

m_{B}= 4500 lbs (FYI, I was wrong, a tank weighs 135,000 lbs)

F-F_{f}= m_{B}a

m_{B}a_{B}- u_{k}m_{B}g= m_{B}a

(4500) (11.17) - .8(4500) (9.8)= 4500a

a= 10.376 m/s

They may not look like a big difference; well, its not in the short distance.

x= 1 mile= 1609.34 m

x= .5at^{2}

t^{2}= 2x/a

t^{2}= 2(1609.34)/ (10.376)

t^{2}= 310.2

t= 17.65 s

vs.

t^{2}= 2x/a

t^{2}= 2(1609.34)/ (11.17)

t^{2}= 287.999

t= 16.97 s

So, obviously, when it comes to short distances, the friction on the car is almost negligible. However, if Superman decided to play a prank on Batman (his attempt to be funny), he could put the Batmobile in space, where the car can infinitely accelerate without friction, where the Batmobile will become the Batrocket. Let's hope the brakes are working.

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