I was driving uphill from a complete stop for a distance of .4 miles estimated to take 1 minute in a navigation app. I was pulled over right after cresting the hill. The cop had me on radar going 53mph. What do I need to know to see if it was possible to be going that fast uphill ? The grade of the incline? The 0 to 60 time capability of my car? I appreciate any help I can get!

# [Physics] Help me get out of a speeding ticket

homework-and-exerciseskinematicsspeed

#### Related Solutions

Alright, let's start with your direct question. Since $d = vt$ the time it takes to travel a certain distance is inversely proportional to your speed $ t \propto v^{-1} $, and so the fractional change in time is proportional to the negative fractional change in your speed.

$$ \frac{dt}{t} = - \frac{dv}{v} $$ So, if we consider typical a typical highway speed of 65 mph and a 5 mph difference, this is a fractional change of about 8%. So going 5 mph faster on the highway will shave 8% off your travel time. So if you had an hour drive, it'll shave off 5 minutes.

## Drag

But. Let's try to consider the added cost of going faster. If you go faster, the wind drag is higher, so your car needs more power to maintain speed. More power means more energy, more energy means more fuel, more fuel means more money.

If we consider just the contribution of wind resistance, we know that $ F \propto v^2 $ for cars, and power is $ P = F v $ so the power consumed by drag goes as $ P \propto v^3 $. Energy consumption is $ E = P t $, so if we consider a drive of fixed length, since $ t \propto v^{-1}$ we have for the contribution of air drag, $ E \propto v^2 $. Now, the energy you get from fuel is proportional to the number of gallons you buy, and the cost scales as the number of gallons so $ \text{fuel cost} \propto v^2 $. So the fractional change in the fuel consumed due to wind drag is: $$ \frac{ d(\text{fuel cost}_{\text{drag}} ) }{ \text{ fuel cost}_{\text{drag}} } = 2 \frac{ dv }{ v } $$ So, for the same 8% increase in speed, you pay an additional 16% in fuel costs due to the loss to air drag.

Naturally air drag isn't the only way we use up fuel to keep a car running, there are all kinds of losses in a car, from inefficiencies in the engine itself, to friction in the various components of the engine, etc. As a simple model, let's say that the power a car consumes is the sum of the air drag and a constant term independent of speed: $P \sim \alpha v^3 + \beta $ for some appropriate choices of $\alpha$ and $\beta$ This would mean that our energy consumption would still be $ E = P t $, so for a constant distance drive, we're talking $$ E \sim \alpha v^2 + \frac{\beta}{v} $$. We can test this model against data from a government study (figure from wikipedia:Fuel_economy_in_automobiles) where for our model, we have $$ \text{mpg} = \frac{ \alpha }{ v^2 + \frac{\beta}{v} } $$

Here I've shown the figure as well as an example fit of our model:

The fit is overlayed in red, and corresponds to $\alpha = 1.5 \times 10^5, \beta = 1.28 \times 10^5 $. Notice that our simple model does pretty well and corresponds to a car that has a highway mph of about 25 mpg. Notice that at high speeds we are seeing the scaling we expect due to air drag alone, for at high speed our operating costs are dominated by air resistance, but it was useful to create the simple model and do the fit in this case because the region of interest is in the overlap region.

## Per hour

Now that we know how the efficiency of our car varies with speed, knowing the average price of gas of $\$3.752$/gallon (from wolfram alpha) we can compute the cost to operate an average car at a given speed:

An in particular, we can compute the additional cost per hour per 5mph increase in speed as a function of speed:

## Per 10 miles

Here I've shown the operating costs as a function of the time spent driving, so as to give costs per hour, which I think is useful for longer drives and something people have a handle on from other areas of life.

If we instead want to look at it as function of the distance travelled, we can look at the car efficiency as the cost to travel 10 miles as a function of speed.

Or we can consider again the change created by a 5 mph increase in speed for a fixed travel distance.

Where here it becomes clear that for a fixed travel distance, as long as you are going less than 40 miles per hour (which for our model was the maximum fuel efficiency speed, and varies per car but the data seems to indicate is about 40 mph across the board), you can always justify speeding by 5 mph from purely economic terms, but at something like highway speeds, it costs you an additional 15 cents or so per 10 miles to go 5 over.

## Traffic Lights

So, up till now we have considered the effectiveness of speeding from an economic perspective in the limit that we are travelling unimpeded down the road. As people have requested in the comments, let's try to figure out how effective speeding is in a more city type environment. This is a difficult problem to address, as traffic lights can have fairly complicated controllers. In particular, in some regions there are Green waves where the lights are designed to allow people travelling at the proper speed to pass unimpeded down long stretches of road. Obviously in this case, you would want to travel at the speed of the green wave and speeding wouldn't help you and would in fact hurt you.

But, sophisticated traffic light controllers are not all that common outside of rich large cities. So, let's try to adopt a *spherical cow* type approximation to traffic lights and assume that traffic lights are independent and just operating on some fixed cycle of green and red. $p$ will be the fraction of time the average traffic light is green, $\tau$ will be the length of a red light, $d$ will be the average distance between traffic lights. If the lights are all operating independently, the distribution of waiting times when we reach a light can be modeled as
$$ P(t) = p \delta(t) + \frac{1}{\tau} ( 1 - p ) \quad 0 \leq t \leq \tau $$ or in words, with probability $p$ we don't have to wait at all, otherwise our waiting time will be uniform up to $\tau$. This distribution has mean and variance
$$ \mu = \frac{\tau}{2} ( 1 - p ) $$
$$ \sigma^2 = \frac{\tau^2}{12} ( 1 - p) ( 3 p + 1 ) $$

Now, if we travel for $N$ blocks, we will have for the average time it takes $$ \langle t \rangle = N \left( \frac{d}{v} + \frac{\tau}{2} ( 1 - p ) \right) $$ $$ \sigma^2_t = N \frac{\tau^2}{12} ( 1- p) ( 3 p + 1 ) $$ where we have added in the travel time between the lights themselves.

So, for instance, with $d = 1/10$ mile between lights on average, $\tau = 30$ seconds, and $p = 0.65$ we get for an average city speed as function of target speed:

So, for a target speed of about 45 mph for a major road in a city, we get for an average speed something like 28 mph, which seems to agree moderately well with observations.

Now, as we have modeled it, if you speed you will get there faster, but what we should compare against is the intrinsic variability introduced by the traffic lights, and a case could be made that speeding 5 mph over is really only worth it if the gains you get in timing are larger than the natural variations in times you would have given the lights, otherwise you'll hardly notice the effect. So in particular, we can compare the fractional reduction in your travel time for going 5 mph over, versus the fractional change in your travel time due to the intrinsic variation due to the random light timings $(\sigma/\mu)$ for different number of blocks. We obtain:

Here the solid line shows the fractional change in your travel time you'd get by going 5 mph over the target speed at the bottom. Notice that it scales as $1/v$ just as the very top of the post. The dotted lines show fractional change in travel time induced by a 1 sigma variation in the behavior of the traffic lights, for different number of blocks. Notice that at around 40 mph, the time you would shave off by going 5 mph is comparable to the natural variations you would expect in travel times due to your luck with the traffic lights if you are travelling 10 blocks, and both of these are at about the 10% level. At this point it starts to become difficult to justify speeding as its effect will be hard to notice over the natural variation. But, notice that if you are travelling a longer distance, there is a clear gain given by speeding, as the variations in travel time start to be suppressed through averaging. On the flip side, for very short trips of a few blocks, the variations in your travel time given by your luck at the lights completely dominates any gain you'd get by speeding.

Here I've simulated traveling for 5, 15 or 50 blocks according to our model, both at 45 mph and going 50 mph. I ran the simulation 10,000 times and here I show the time gained from speeding in different trials. Here we can really see that there are no noticeable gains for going 5 over over 5 blocks, it's completely washed out by our luck with the lights, but a noticeable change in our expected time over 50 blocks.

The code used for this answer is available as an ipython notebook

*Addendum:* I've explore a more realistic model of power losses in cars in this more recent answer

The tricky part of this question is that you are given a graph of *velocity* but asked a question about *speed*.

Several others have said essentially the same thing, but what really makes this clear for me is a graph of speed:

The above is the graph of $$ y = \left \lvert 4 - \left ( \frac{x - 2}{2} \right ) ^2 \right \rvert \text{,}$$ which is just the absolute value of the velocity graph in your screenshot.

This represents the fact that **speed is the absolute value of velocity**.

We understand "slowing down" to mean that the slope of the speed is negative, and "speeding up" to mean that the slope of the speed is positive. What is the slope of point $(6, 0)$ on the graph (which corresponds to your circled dot)?

This point is a cusp. The notion of "slope" only exists for differentiable points, and as Wikipedia says,

a function with a bend, cusp, or vertical tangent may be continuous, but fails to be differentiable at the location of the anomaly.

Thus *the slope of speed does not exist at this point*, and so the object is neither speeding up nor slowing down in this instant.

## Best Answer

The best way to solve it would be experimentally, by doing the run several times, with calibrated instrumentation by the roadside to measure your speed. The acceleration won't have been constant, so that's not an assumption we can use.

Knowing the 0-60 time capability won't really help; it could be different when accelerating up hill, compared to on the flat. You'll get a boost at the very crest of the hill, as your car goes from accelerating up hill to accelerating on the flat. So the measured speed could be very sensitive to the place at which your speed is measured.

Acceleration won't have been uniform: all you know is that the second integral of the acceleration is 0.4 miles.

Your 1-minute estimate of time taken isn't anywhere near precise enough to be useful - what's the uncertainty on that, $\pm25\% $ ? And the distance estimate of 0.4 miles, assuming 1 s.f., would appear to have an uncertainty of $\pm13\%$.

All in all, there are so many complicating factors, that you will need to run the experiment to get to the bottom of it. This could turn out to be an expensive question to answer.

The question does illustrate the importance of assessing uncertainty in your inputs, and questioning assumptions (such as uniform acceleration) that would make calculation easy, but are unrealistic, and thus would give an invalid answer..