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I am looking for a formula to estimate how fast a vehicle travels down a dirt track with the following data:

Jeep #$1$: $1000$ hp, weighs $1700$ pounds and travels $300$ ft. in $3.6$ seconds

Jeep #$2$: $600$ hp, weighs $2500$ pounds and travels $300$ ft. in $4.9$ seconds

Blazer #$1$: $750$ hp, weighs $3000$ pounds and travels $300$ ft. in $4.7$ seconds

Blazer #$2$ $1000$ hp, weighs $2700$ pounds and travels $300$ ft. in how many seconds?

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    +1 @Aryabhata. Also, three data points are pitifully little with which to develop any kind of model. A reasonable model would have at least two adjustable parameters (encoding the effect of power and weight, respectively), and at least one of these parameters must be different between the two vehicle types. Already, then, there are so many parameters that _every_ set of 3 given observations can be explained by fiddling with them, and we have no way to see whether the _structure_ of the model fits the data or not.2012-02-28

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It's a long shot but i'll give it a try. Let's assume that the average speed of the jeep and the blazer is propotional to the horse-power and the weight. So we will have
$Speed=a*HP-b*W$ Using the 2 sets of data we have for the jeep,we build the following equation system:
$1. \space \space \space\;\;\;(300/3.6)=83.33=a1000-b1700$ $2. \space \space \space\;\;\;(300/4.9)=61.22=a600-b2500$ By solving the above system we obtain $a_{jeep}=0.07$ and $b_{jeep}=0.00768$. We assume that the effect of the weight in the overall performance is the same for the Blazer so $b_{blazer}=0.00768$. From the Blazer #1 we obtain that $a_{blazer}=0.1158$.Finally we have for the 4th car:
$speed=a*HP-b*W=0.1158*1000-0.00768*2700=95.09 \Leftrightarrow time=3.15s$

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    @Henning : Enjoy the world of mathematics, where heavy cars break the entire house instead of leaving the garage from the open garage door exit.2012-03-30