Transportation
Energy Use in Cars 2: Constant Speed Cruising
Problem Set Solutions
Problem 1: People are also about 25% efficient at turning chemical energy into mechanical energy. If
the CD of a bicycle is 0.9 and a bicyclist travels at an average of 20 km/h, estimate the fuel cost in
kWh/km of bicycle travel at constant speed. Compare this to a car travelling at 50 km/h. (You can
neglect rolling resistance).
Solution 1:
Approach: Calculate the work done against air resistance for each kilometre the bicycle travels. Use the
efficiency to find out how much fuel you need. Convert fuel cost to kWh/km. Compare to fuel cost of a
car (calculated in Energy Use in Cars 2).
What we know (bicycle):
CD = 0.9
v = 20 km/h = 5.5 m/s
air= 1.3 kg/m3
A ~ 0.5 m x 1.5 m = 1.75 m2 (assume bicycle is a little less wide than a car)
d = 1 km
Calculating work done against air resistance per kilometre for the bicycle:
1
Work  AC D dv 2
2
1
 (1.3kg / m 3 )(1.75m 2 )(0.9)(1000m)(5.5m / s) 2
2
 31kJ
So, 31 kJ of work is done against air resistance for each kilometre the bicycle travels.
Calculate how much fuel is required by taking into account the efficiency:
WorkOutput
Efficiency
31kJ

25%
 124kJ
Fuel Energy Input 
For each kilometre, 124 kJ of fuel is required.
Physics and Astronomy Outreach Program at the University of British Columbia
Convert result into KWh/km:
1kJ  (1kW  s )
1h
3600s
3600kJ  1kW  h
124kJ 
1kW  h
 0.034kW  h
3600kJ
Similarly for a car:
From Energy Use in Cars 2 the work done against air resistance for a car travelling at 50 km/h is 505.4 kJ
for each kilometre.
Convert to kWh/km:
1kJ  (1kW  s )
1hr
3600s
3600kJ  1kW  h
505.4kJ 
1kW  h
 0.14kW  h / km
3600kJ
Comparison:
fuel of bicycle 0.034

 25%
fuel of car
0.14
Comparing these two values we find that the fuel cost of a bicycle travelling at 20 km/h is about 25%

that of a car travelling
at 50 km/h.
Mathew (Sandy) Martinuk 2009/08/25
Physics and Astronomy Outreach Program at the University of British Columbia