A 4500-lbm truck approaches a stop sign at 20 mph. (a) What is the kinetic energ

Question

A 4500-lbm truck approaches a stop sign at 20 mph.
(a) What is the kinetic energy of the vehicle in Btu?
(b) If the maximum energy content of gasoline is 112,000 Btu/lbm and the specific gravity is
SG=0.7, how many gallons of fuel are saved by not stopping at the stop sign?
(c) If the actual efficiency of the truck's engine is 15% at this low velocity, how many gallons of fuel
are actually saved by not stopping at the stop sign?
(d) If gasoline costs $4.10/gallon, how much money is saved by not stopping at the stop sign? Use
your answer from part (c).
(e) If the traffic ticket for running a stop sign is $65, at what probability (or percent chance) of
getting a ticket makes it economically advantageous to not stop at the stop sign?

Please include clear work with all steps. Explain all formulas used and show what the variables mean (not everyone uses the same variables). First clear work gets a lifesaver. Answers are here, but I need the steps:

(a)77.3Btu
(b)0.0001188 gal
(c)0.0007918 gal
(d)$0.00324
(e)0.00005 or a 1 in 20,000 chance.

Explanation / Answer

(a)

m = 0.453592*4500 kg

v=20 mph =8.940 m/s

1055 J per BTU

E = 0.5*m*v^2 = (0.5*4500*20*8.94^2)/1055 = 77.3 Btu

(b)

Pound of gas = 77.3/112000

Litre of gas = 77.3*0.453592/112000*0.7

= 4.4722278*10^-4 l

4 l =1 gal

4.4722278*10^-4 l = 0.0001188 gal

(c)

gallons of fuel =0.15*0.0001188 = 0.0007918 gal

(d)

Money saved = 4.1*0.0007918 = $0.00324

(e)

chance = 65/0.00324

= 20061

P =1/20061 = 0.0000498 = 0.00005

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