- Asked By Judahjohnston
- Asked on December 21, 2009

- Reopen Abandoned Water Contamination Investigations
- Save Our Oceans from Destructive Overfishing
- Demand an End to Wolf-Killing Competition
- Save Turtles from Extinction
- Stop Luxury Cruise from Destroying Arctic Glaciers
- Applaud Travel Company for Banning Elephant Tours
- Protect Arctic Wildlife from Oil Drilling
- Protect Migrating Birds from Invisible-Glass Football Stadium
- Stop Fur Farming
- Thank Tesla for Making Electric Vehicles More Accessible

1

## Answers

There are a lot of variables at play here. Most importantly— what is the weight of the car? A smart car (tiny) obviously requires a lot less energy to get going 60 miles per hour than a limosine, a hummer, or an 18 wheeler. But let’s use some averages and try and figure out an estimate:

Car’s weight: 3500 lbs

Speed: 60 miles per hour

Efficiency of engine/drivetrain: 15 percent

We have to use some physics equations:

KE = 1/2 mv^2

We need to convert (this is why the english standard system is weak)

3500 lbs x (1 kg / 2.2 lbs) = 1587.573 kg

and 60 mph to meters per second = 96560.64 m / 3600 seconds = 26.8224 m/s

OK now we know how much energy the thing will have = .5 x 1587.573 x (26.8224 ^2) = .5 x 1587.573 x 719.44114 = 571,082.66 joules of energy — but that’s how much actually got to the drive train — remember this number is only 15 percent of the total energy used: so we will use another formula:

.15 x T (the total amount of energy the engine burned up from the fuel) = 571,082.66

T = 571,082.66 / .15 = 3,807,217.7 joules which converts to about a kilowatt hour of energy

or exactly 1.057560472 kWh !

Any physics experts out there willing to check my work?

KE=.5 x mass x (speed squared)