Here's the context:

Due to roadwork ahead, the traffic on a highway slows linearly from 55 mph to 15 mph over a 2000-foot stretch of road, then crawls along at 15 mph for 5000 feet, then speeds back up linearly in the next 1000 feet, after which it moves steadily at 55 mph.

I've already answered parts a-e, which asked for the piece wise function representation of a vector field in miles/hour, the divergence at multiple points, a conceptual question on comparing densities of traffic, and a few questions about density times the vector field.

The next couple parts are ones that I am confused about:

f. Determine p(x) if p(0)=75 and pv is constant. It gives the hint that p(x)=4125/(55-x/50) when 0<=x<=2000

e. If the highway has two lanes, find the approximate number of feet between cars at x = 0, x = 1000 and x = 5000.

Could someone please help ?

  • $\begingroup$ As a scientific person, why don't you use metric system ? $\endgroup$
    – Jean Marie
    Dec 4, 2017 at 8:10

1 Answer 1


For point f: if $pv$ is constant, since at time zero $p(0)=75\,\,$ and $v(0)=55\,\,$, then at each instant we must have

$$p(x) \cdot v(x)=75 \cdot 55=4125$$

Now in the tract $0 \leq x \leq 2000\,\,$ we have that $v$ linearly decreases from $55$ to $15$ over $2000$ feet, which means that the rate of slowing is $40/2000=1/50\,\,$ per foot. Thus, after $x$ feet, the velocity is reduced by $x/50$. The equation above then becomes

$$p(x) \left(55-\frac{x}{50} \right)=4125$$

$$p(x) =\frac{4125}{ \left(55-\frac{x}{50} \right) }$$

Continuing in this way, you can easily determine $p(x)$ for the whole path.

  • $\begingroup$ For the second part, would I just use p(x)⋅v(x) and convert it into cars/feet and use that to determine feet between cars? $\endgroup$
    – Melanie
    Dec 4, 2017 at 18:22
  • $\begingroup$ Yes, it could be a good approach. $\endgroup$
    – Anatoly
    Dec 4, 2017 at 22:27

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