I think you're saying that the speed at position x is 1-x. (Where the initial point is 0 and the endpoint is at 1). So the time at position x is integral dx/(1-x) from 0 to x, or -ln(1-x). Solving for x in terms of t gives x = 1-exp(-t), and indeed v = dx/dt = exp(-t) = 1-x. Acceleration is dv/dt = -exp(-t) = x-1. Amusing. --ms On 6/5/2012 1:35 PM, Simon Plouffe wrote:
Hello,
I was going back from a place about 100 km away from home and at one point the GPS I have was kind of stuck on a fix time of arrival, which later was corrected I presume.
But it made me think of a problem which is the following.
Suppose you are at exactly 100 Km from a place going on a straight road at a speed of 100 Km/hour. Obviously, it will take you 1 hour exactly to reach your destination.
BUT : what would be your increase in speed or decrease IF , let's say you make a point every kilometer you made so far so that : It would be always 1 hour in time to reach your destination ?
In other words, what is the change in speed IF at every kilometer there is always 1 hour left of driving until you reach destination.
After 1 min at 100 km/hour, there is 99 kilometers left so if you need to reach destination in 1 hour you need to go at 99 km/hour.
Is this a trivial problem ? My impression is that the speed would necessarly decrease to near 0 when you are 1 mm away from your home.
In plain english , what is the function of the speed ?
If someone has a solution, I would be glad to ear it.
Best regards,
Simon plouffe
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