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Part A選擇題部分:
Part B簡答題部分
Problem 1
(a) To compare the e?ciencies, we need to convert them to the same unit:
3.785 L/gal
USA car e?ciency = 30 mpg ? =7.84 L/100km
(30 mpg)×(1.609 km/mile)
European car e?ciency = 7.80 L/100km
The European car is the most e?cient.
(b)
0.2 × 7.8 L/100km =1.56 L/100km ? European car new e?ciency = 6.24 L/100km
0.2 × 30 mpg =6 mpg ? USA car new e?ciency = 36 mpg
3.785 L/gal
USA car new e?ciency = 36 mpg ? =6.53 L/100km
(36 mpg)×(1.609 km/mile)
Therefore the European car is still the most e?cient, by 6.53?6.24 ×100% = 4.65%. Note that a 20% improvement
6.24
in e?ciency measured in L/100km is more important than a 20% improvement in e?ciency measured in mpg.
(c) The distance traveled in each part of the trip is given by the velocity during this part, multiplied by the travel time, so the total distance traveled is:
ΔX =0.5h × 60km/h +0.5h × 120km/h +0.5h × 80km/h +0.5h × 100km/h = 180km
(d) The work done by the the drag force is the sum of the work in each part of the motion:
4
W =i=1 FiΔXi = FiViΔti = 240N ×60km/h ×0.5h+540N ×120km/h ×0.5h+320N ×80km/h ×0.5h +420N ×100km/h ×0.5h = 73.4 MJ
(e) The total gasoline used is given by the sum of the amounts used in each part of the motion:
4
FiViΔti
l ==
i=1 35 MJ/L×E?ciencyi 240N×60km/h 540N×120km/h 320N×80km/h 420N×100km/h
0.5h × (+ ++)/(35 MJ/L) = 16.7 L
0.075 0.15 0.11 0.135
(f) The amount of gasoline used per kilometer travelled (denoted by lkm) for each part of the motion is then given by:
FV ΔtF 1 J/m F 1 L
== × = ×× 10?3
lkm 35MJ/L×E?ciency×V Δt E?ciency 35 MJ/L E?ciency 35 km
For the di?erent speeds of the car the lkm can be calculated:
V = 60km/h ? lkm =9.14 L/100km V = 120km/h ? lkm = 10.3 L/100km V = 80km/h ? lkm =8.31 L/100km V = 100km/h ? lkm =8.89 L/100km
Therefore gasoline is used most e?ciently when the car is traveling at 80km/h .
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