# Answered on Quora: What burns more gas, accelerating as fast as possible to 60 mph (e.g. 10 seconds) or accelerating slowly (e.g. 30 seconds)?

What burns more gas, accelerating as fast as possible to 60 mph (e.g. 10 seconds) or accelerating slowly (e.g. 30 seconds)?

Daniel Becker, Patent Attorney and Early-stage Engineering (2013-present)

strict answer: hard acceleration and then coasting

VERY oversimplistic explanation: power is related to fuel consumption rate, and acceleration fuel usage is several multiples of the rate of coasting. So driving smoothly and accelerating gently means spending more time at some percentage of maximum fuel consumption, rather than spending the most time at the dramatically lower fuel usage rate. I’l just make up some hopefully reasonable numbers, to illustrate.

A fit might require 15 hp for highway cruising, but accelerating gently means average 45, and maximum acceleration requires average 90 hp. Let’s just go ahead and imagine translating those, 90 hp 1/6th of the fuel efficiency of 15 hp, and 45 hp 1/3 the fuel consumption of 15 hp (presumes a best of all worlds torque curve, btw).

45 mpg cruising at desired speed (presume 60 mph)

0–60 at 45 hp average… est 20 seconds

0–60 at 90 hp average … est 10 seconds

distance to be covered, minimum distance necessary to read the desired speed at 45 hp average =

v^2=v0^2+2ax

60^2 mph^2 =2 a (mph/s) x (m)

3600 mph^2 = 2 * (60 mph/20 s) *x (miles)

60 mph = x miles/20s

1200 mph *s = x miles

1200 / 3600 m/s *s = 1/3 mile

so, 0 s spent at 45 mpg

1/3 mile * 1 gallon / 15 miles = 1/45th of a gallon = 0.022 gallons

now… we need to know how far was traveled over the 10 seconds of hard acceleration

v^2=v0^2 +2 ax

x= v^2/(2*a)

x= (60 ^2 mph^2) / (2 * 60 mph/10 s)

x= 300 mph*s

x= 1/12 (miles/second)*s = 1/12 mile

1/3 = 4/12 miles to accelerate gently

vs.

1/12 miles to accelerate hard , 4/12- 1/12= 3/12 remaining= 1/4 miles coasting, after accelerating hard

(1/12 miles * 1/(45 mpg *1/6 efficiency))+ (1/4 miles *1/ 45 mpg)

= 1/(12*7.5) + (1/(4*45))= (1/90) + (1/180)= 3/180 = 1/60 = 0.0166 g

so, over 1/3 mile=

gentle = 0.022 gallons, or 15 mpg

hard+coast= 0.0166, 19.88 mpg.

4.88/15= 32.5% better fuel economy during a distance over which gentle acceleration requires, to reach cruising velocity (b/c presuming the coasting is the same consumption for either driving style, after that distance).

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