# Simpson's rule to estimate distance traveled given velocity at certain points

Problem:

A boat drives a steady course with a variable speed for 4 hours. The speed is registered at regular intervals in meters per second.

The registration shows $2.4, 4.4, 7.6, 8.4, 8.6, 7.9, 8.3, 8.7, 7.7, 6.5, 7.1, 6.7, 1.4$ (sorry, but I'm making a point out of this later).

Use Simpson's Rule to estimate how far the boat has traveled during the four hours, and its average speed.

Additional question; why can Simpson's Rule be used to estimate the boat's distance traveled?

My progress:

There are 13 recordings, which is an odd number, so that should be fine.

Using the Composite Simpson's Rule (with coefficients 1,4,2,4,2...,2,4,1) , I get $$\frac19\left[ 2.4 + 4(4.4) + 2(7.6) + 4(8.4) + \ldots + 4(6.7) + 1.4 \right] \approx 28.09$$ (unless I've made some careless mistake).

Here $\frac19$ comes from $\frac h3$ (from formula) where $h = \frac13$.

Now, as I understand it, this number would - since we're estimating the area under the "curve" (i.e. using Simpson to estimate a definite integral given points) - be a number for the distance traveled.

But since the numbers are given in $\frac ms$, wouldn't this $28.09m$ be an oddly low number? What am I missing here?

Any help appreciated!

I believe your $\frac{1}{9}$ is obtained by $\frac{1}{3}\cdot\frac{1}{3}$ where one of the $\frac{1}{3}$ is in hours. So instead you should multiply by $3600$, which is the number of seconds in an hour.

• I have no answer to check by, but I end up with around 101km, which makes more sense than 28m.
– Alec
May 12, 2015 at 20:42

You did not give the details about how you got the coefficient $\frac 19$, but you appear to be integrating the speed over the interval $[0,4]$. That's OK if all your times are measured in hours, but as you noted, the speeds are all measured in meters per second.

Use consistent units throughout the application of Simpson's Rule. Either convert all the speeds to meters per hour, or better, convert $4$ hours to $t$ seconds and integrate over the interval $[0,t]$.

• Sorry, edited in the part about $\frac19$.
– Alec
May 12, 2015 at 18:37
• So $h$ is $\frac13$ hours? Then it is exactly as the answer by KittyL says, and you should use $h = 1200$ (measured in seconds) instead. In any case, what you're doing is equivalent to integration over the interval $[0,12h]$, so if your speeds are meters/second, $12h$ needs to be equal to the number of seconds the boat was motoring, not $4$ seconds. May 12, 2015 at 19:33

I have calculated the answer using simpson's one-third rule(the distance travelled by boat in 4 hours) 93.56307km.

• This answer is correct as per simpson's one third rule. May 14, 2016 at 4:52