average speed calculation - Date diff help

Hi,

I need to work out the average speed which is calculated as
Average Speed = distance ÷ time

I know the following
Start Time = '2014-09-12 07:03:36.000'
End Time = '2014-09-12 07:06:22.000'
Distance = 6km

So i need to figure out the SQL which works out the TIME as a float:

I'm having issues figuring out the Time as a float (i.e x.x),
 if i use

declare @startTime datetime = '2014-09-12 07:03:36.000'
declare @endTime datetime = '2014-09-12 07:06:22.000'
print convert(float, DATEDIFF(minute,@startTime, @endTime))

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This outputs 3   (minutes - but rounded up)

If i change MINUTE to SECOND

declare @startTime datetime = '2014-09-12 07:03:36.000'
declare @endTime datetime = '2014-09-12 07:06:22.000'
print convert(float, DATEDIFF(SECOND,@startTime, @endTime))

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This Outputs 166 (seconds)

if I try both these calculations to work out average speed: (distance ÷ time):

print 6 / 166   -- this equals 0 (second variation)
print 6 / 3       -- this equals 2 (minute variation)

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The problem is the MINUTE variation doesn't take into account seconds, so its rounded UP to 3 mins and its not accurate, and the SECONDS variation just doesn't work

I guess i need to use a combination of both, and some how get the remaining seconds and use as a point
i.e.

Print 6 / 2.8 

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however i've no idea how to do this

I guess i need to use the SECOND variation which outputs 166 (seconds) and turn that into a minute variation of 2.X ?

HELP