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Originally Posted by Carrameow
My defense is based on the fact that the police car was a 1/2 mile from where i was stopped at the traffic light and where he spotted me. So if say my 0 to 60 time is 15 seconds, my acceleration is (60mph)/(15 sec), or .0011111 miles/sec squared (there are 3600 sec in an hour). Then i could calculate the fastest I could go in that 1/2 mile using vexp2= 2ax , but my physics and my head are mushy, and somehow I just got my half mile time as 100 mph, which I think is wrong.
What I am really trying to find is a 1/2 mile time for a 300D, whats the highest speed I could do from dead stop to a 1/2 mile at full acceleration...
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I think you are wasting you time.
But, here goes:
Let's work in feet/second, which is easier.
The acceleration of the 300D is 6 ft./secē (0-60 in 15 sec.)
The distance that you are interested in is 2640 feet.
The formula for velocity is: Vē=2AS. A is acceleration, S is distance.
So your velocity at 1/2 mile is 178 ft./sec. or 121 mph.
HOWEVER: The 300D cannot sustain an acceleration of 6 ft./secē between the speeds of 60 mph and 121 mph. This is due to the very strong effect of wind resistance on the body. The vehicle would never reach 121 mph within 1/2 mile.
Hell, it wouldn't reach 121 mph in 5 miles.
The actual answer to your question can only be acheived with a stopwatch.
