Quote:
Originally Posted by ah-kay
1466.67' is correct but it has NOTHING to do acceleration. The terminal speed and the time taken is enough to calculate the distance travel.
The average speed is (100-0)/2 = 50mph. How it gets there is not important. 50mph for 20 sec = ((50*1760*3)/3600)*20 = 1466.67'
BTW: I assume it is mph.
|
Not true !!
You have not calculated the average speed !!
If we assume constant acceleration, we can use one of the equations Botnst links to ;
Distance = initial velocity X time + 1/2 X acceleration X time squared.
If acceleration varies, the area under a velocity/ time graph would give you the answer.
BC is correct in converting to compatible units.
velocity in ft/sec , time in sec , distance in ft.
__________________
Grumpy Old Diesel Owners Club group
I no longer question authority, I annoy authority. More effect, less effort....
1967 230-6 auto parts car. rust bucket.
1980 300D now parts car 800k miles
1984 300D 500k miles
1987 250td 160k miles English import
2001 jeep turbo diesel 130k miles
1998 jeep tdi ~ followed me home. Needs a turbo.
1968 Ford F750 truck. 6-354 diesel conversion.
Other toys ~J.D.,Cat & GM ~ mainly earth moving