Lab 12: Example (distance-time-velocity)

In Lab 7, we were given information about the NASA Q36 Robotic Lunar Rover. Specifically, it can travel up to 3 hours on a single charge and has a range of 1.6 miles. After \(t\) hours of traveling, its speed is \(v(t)\) miles per hour given by the function \(v(t) = \sin\sqrt{9 - t^2}\) . One hour into a trip, the Q36 will have traveled 0.19655 miles. Two hours into a trip, the Q36 will have traveled 0.72421 miles. Now we’ll see how to calculate values for the distance traveled as a function of time, even though we won’t be able to write that function in terms of elementary functions.

In Lab 7, we computed the acceleration \(\displaystyle a(t) = v'(t) = \frac{-t \cos\sqrt{9 - t^2} }{\sqrt{9 -t^2} }\) which is positive from \(t = 0\) to \(t = \sqrt{9 - \pi^2/4} \approx 2.556 \).

Let’s approximate the distance traveled by the Q36 in the first two hours.


A First Approximation

We could get quick, but not very accurate, approximations by pretending that the Q36 traveled at a constant speed the entire time. Since the rover is always speeding up during these two hours, using the initial and final speeds would produce an underestimate and overestimate for the distance traveled, respectively:

Underestimate: 2 hours at \(v(0) = 0.14112\) mph is 0.28224 miles. Overestimate: 2 hours at \(v(2) = 0.78675\) mph is 1.57350 miles.

This is a huge range, so using either of these as an approximation, the best we can say is that we are within 1.57350 – 0.28224 = 1.29126 miles of the exact answer. Of course we were told that the rover travels 0.72421 miles in the first two hours, so in this case we can compute the exact errors (usually not possible, otherwise you wouldn’t be approximating):

The underestimate 0.28224 miles is \(|0.28224 – 0.72421| = 0.44197\) miles off.

The overestimate 1.57350 miles is \(|1.57350 – 0.72421| = 0.84929\) miles off.

Although not very accurate, both of these errors are smaller than our computed error bound: \[ \text{error} < 1.29126 \text{ miles}.\] Neither of these approximations is exact because the Q36 is not traveling at a constant speed, thus simply using \(d = vt\) isn’t sufficient.


A Better Approximation

Since we can compute the velocity at any time we want, we could get a better approximation by assuming a constant speed but using shorter time intervals.

Let’s break the 2 hours into four \(\frac{1}{2}\)-hour subintervals, that is \(\Delta t = \frac{1}{2}\). We can still use the fact that the Q36 is speeding up to get underestimates and overestimates:

We’ll use the velocity at the beginning of each \(\frac{1}{2}\)-hour to get underestimates.

We’ll use the velocity at the end of each \(\frac{1}{2}\)-hour to get overestimates.

Consider the following table of velocities:

Time \(t\) in hours 0 0.5 1 1.5 2
Velocity \(v(t)\) in mph 0.14112 0.18252 0.30807 0.51715 0.78675

Assuming the speed at the beginning of each half hour, we would determine the Q36 traveled \[ \frac{1}{2}(0.14112 ) + \frac{1}{2}(0.18252) + \frac{1}{2}(0.30807) + \frac{1}{2}(0.51715) = 0.57443 \text{ miles}.\] Assuming the speed at the end of each half hour, we would determine the Q36 traveled \[ \frac{1}{2}(0.18252 ) + \frac{1}{2}(0.30807) + \frac{1}{2}(0.51715) + \frac{1}{2}(0.78675) = 0.89725 \text{ miles}.\] For this underestimate/overestimate pair, the error bound is \(0.89725 - 0.57443 = 0.32282\) miles. So this is significantly more accurate than our first attempt, but these approximations may still be nearly \(\frac{1}{3}\)-mile off.



Refining the Approximations

Now let’s break the two hours into twelve 10-minute subintervals, or t = 1/6 hour. To get both an underestimate and an overestimate, we’ll need the following 13 velocity values:

Time \(t\) in hours 0 1/6 1/3 1/2 2/3 5/6 1
Velocity \(v(t)\) in mph 0.14112 0.14571 0.15948 0.18252 0.21491 0.25675 0.30807


Time \(t\) in hours 7/6 4/3 3/2 5/3 11/6 2
Velocity \(v(t)\) in mph 0.36882 0.43872 0.51715 0.60292 0.69395 0.78675


Assuming the speed at the beginning of each 10 minutes, we underestimate the distance traveled by the rover is 0.67169 miles. Assuming the speed at the end of each 10 minutes, we overestimate the distance is 0.77929 miles. We can bound the error for these approximations by their difference, of 0.1076 miles.

Notice that in computing the bound on the error by subtracting the overestimate minus the underestimate, most of the terms cancel. We can use this to determine error bounds before we even start to find approximations! In general, the error bound will be \(0.78675 \Delta t – 0.14112 \Delta t\), or simplifying, we can say error \( < 0.64563 \Delta t\).