A car is moving down a street with no brakes or gas. The car is slowing due to wind resistance and the effect of friction. The road is flat and straight. The only data I have are timings taken at 100m intervals along its path. I have the following measurements:

Marker (100m intervals)     Time (seconds)
Start                       N/A
100m                        6.49     
200m                        7.66
300m                        9.15
400m                        10.71

Is this enough information to calculate how much time until the car reaches a specific speed of 2 m/sec?

What I'm Trying:

If I could get the deceleration rate, I could use: $$\ D=vt+(1/2)at^2 $$

I can solve for v to get a velocity, and then use the velocity and deceleration to figure out when the car will reach 2m/s.

But the deceleration is not constant. The best I can do is get average deceleration over an interval. It seems like there should be a way to get the deceleration curve based only on time across a distance.

I'm lost.. any help is greatly appreciated.

  • $\begingroup$ The chart is a bit ambiguous. On first read, it looks like the car is accelerating. But, I suppose, the "Time" is the elapsed time between markers rather than the elapsed time from start? Anyhow, I think I would approach this by plotting the the elapsed time from "Start" and estimate the limiting value of the deceleration. As the speed falls, the wind resistance becomes insignificant compared to the friction so, after some point, the deceleration is approximately constant. This should allow you to estimate the time to reach 2 m/sec. $\endgroup$ Jul 17, 2012 at 1:46
  • $\begingroup$ Is the time at the start really N/A, or is it actually zero seconds? $\endgroup$
    – David Z
    Jul 17, 2012 at 2:10
  • $\begingroup$ The car is already moving for some amount of time before it moves across the markers. Alfred is correct in assuming that the time column is the amount of time from one marker to the next. The first time is N/A because its the start of the timer. Sorry for the confusion, I could have spelled that out better. $\endgroup$
    – Jeff
    Jul 17, 2012 at 3:03
  • $\begingroup$ So Alfred, you are saying that in the absence of air resistance, the deceleration should be fairly constant. That helps. I'm still unsure how to get deceleration and the velocities at any specific point in time. Any thoughts? $\endgroup$
    – Jeff
    Jul 17, 2012 at 3:12

1 Answer 1


I can't see a good way to answer your question without some more data.

I tried fitting a quadratic and cubic to your distance time data, and both give pretty good fits, but the fitted velocity at the 400m point is around 8m/sec so it's a big extrapolation down to 2m/sec. In fact the quadratic and cubic fits diverge quite strongly beyond about 450m so I wouldn't trust either of them.

I think the only good way to estimate the time to 2m/sec is if you know a specific mathematical model underlies the motion. If this is coursework there may be some hints about what model they expect you to use. If it's a real life example I think you'd be bold to assume any particular model. As I recall, resistance tends to go as $v^2$ at high speeds and switch to $v$ at low speeds, with an additional constant (i.e. independant of velocity) resistance due to friction in the drive train. But the details will depend on the car.

If you want the fits (fitted using Excel) they are:

$$ s = -0.1148t^2 + 15.596t + 1.4563 $$ $$ s = 0.0017t^3 - 0.201t^2 + 16.64t + 0.0021 $$

Later: I'm probably taking this further than the data supports, but if you assume the deceleration is proportional to $v^2$ you get $v \propto 1/t$ and hence $s \propto ln(t)$. If I fit a function of this form to your data I find the following gives a pretty good fit:

$$ s = 577.5 \space ln(t + 34.1) + 2038 $$

The point of this fit is that it's physically motivated while the fits above were purely numerical. Anyhow, differentiating this to get the velocity gives:

$$ v = \frac{577.5}{t + 34.1} $$

So if you're prepared to assume the deceleration stays proportional to $v^2$, the car reaches 2 m/sec after 255 seconds.

Response to comment: I can only speak as an (ex) experimental physicist, but faced with some data the first thing I would do is graph it and see what the curve looks like. Experience often immediately tells you what the functional form is e.g. whether it's a parabola, hyperbola, exponential etc. However in this case nothing obvious sprang to mind.

The next step is to fit a few random functions and see if anything looks plausible. Excel has a polynomial fit built in. Graph your data as a scatter plot then right click the curve in the graph and select "add trendline". Select the type of fit, then on the options tab click "include formula" or you'll just get the line and not the fitted equation.

In this case the polynomial fit didn't shed any light, so you have to get out the pencil and paper and try to work out a physically reasonable description. Wikipedia will tell you that at high speed air resistance goes as $v^2$, so a reasonable guess is that the retarding force is proportional to $v^2$. I used my polynomial fits to calculate $v$ and $a$ from your data. With only five points the results were pretty noisy, but it did look as if a 2-fold change in velocity did cause a 4-fold change in acceleration so the $v^2$ dependance looked plausible.

So the equation of motion is something like:

$$ \frac{dv}{dt} = -A \space v^2 $$

for some constant, $A$, and the solution to this is:

$$ v = \frac{B}{t} $$

for some other constant $B$, i.e. velocity falls of as 1/time. This function for $v$ says that $v$ takes infinite time to fall to zero, but we expect it only applies at high speeds so that's OK. Since $v = ds/dt$ we integrate the expression for $v$ to give the logarithmic dependance of distance on time.

The equation I used:

$$ s = a \space ln(t + b) + c $$

is just the most general form of the equation for $s$. I went back to excel and graphed the data along with an extra column for the log equation, then I manually adjusted the values of $a$, $b$ and $c$ to get a rough fit.

At this point I used an Excel add-on called the Solver. I won't go into how to use this because it's quite fiddly and there are lots of articles on how to use it out there in Googleland. The solver can adjust variables you specify to get the best fit of your function to data, and that gave me the equation I posted.

On last comment, extrapolating data is always dangerous because you have to assume that whatever equation you use applies outside the range of data you've collected. The velocity in your data falls from about 16 m/sec at $t = 0$ to about 8 m/sec at the last point in your data, and it's a large extrapolation to try and extend this to 2 m/sec. I doubt that the $v^2$ dependance of deceleration would apply down to 2 m/sec so my calculated value for the time taken to decelerate to 2 m/sec is probably much too high.

  • $\begingroup$ Thanks for your work on this. Can I ask you how you went about fitting the function? Or how I could go about finding a function that fit? These were real world measurements that were taken, but I would like to be able to find a way to create an equation that would represent the deceleration depending on the situation. $\endgroup$
    – Jeff
    Jul 18, 2012 at 5:51
  • $\begingroup$ This is great. Thank you for your help. I'm going to work through this a bit on my own using the tools you've mentioned. Thanks for the detailed response. $\endgroup$
    – Jeff
    Jul 23, 2012 at 20:02

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