The table shows position of a cyclist. t(seconds):0,1,2,3,4,5. s(meters):0,1.4,5.3,10.5,17.4,25.1 Find average velocity for time periods: [1,3] [2,3] [3,5] [3,4].Find instantaneous velocity when t=3

Answer 1

Since velocity is the slope of the distance vs time graph, average velocity is [f(x₂) - f(x₁)]/(x₂ - x₁), or, is this case, [s(t₂) - s(t₁)]/(t₂ - t₁) so for:

t = [1, 3], v_avg = [s(3) - s(1)]/(3 - 1) = (10.5 - 1.4)/2 = 4.55 m/s

t = [2, 3], v_avg = [s(3) - s(2)]/(3 - 2) = (10.5 - 5.3)/1 = 5.2 m/s

t = [3, 5], v_avg = [s(5) - s(3)]/(5 - 3) = (25.1 - 10.5)/2 = 7.3 m/s

t = [3, 4], v_avg = [s(4) - s(3)]/(4 - 3) = (17.4 - 10.5)/1 = 6.9 m/s

The instantaneous velocity at a point "t" is the limit (as h→0) of [s(t + h) - s(t)]/h (so pick smaller and smaller intervals of time to find the "average velocity" until the interval is so small it becomes "instantaneous".

However, to do this with a table of values is a bit of a "contest of opinion" because you do not know what the values of "s" are for "sub-interval" t values, for example, if I want to pick 0.1 for h, I do not know the "s value" at time 3.1 s. So, you must "guess" in some way.

Typically, a straight line approximation often is used to approximate the interim values. I, personally, like to curve fit using the data points and for good curve fits (r² values close to 1) I like to use those functions to estimate the values at interim "t" values. I'll do both here:

We've already calculated the slope of the straight line between 3 and 4 above, it is 6.9. Since it is a straight line, the slope will always be 6.9, so we don't need to use the limit equation. But, to get a better approximation, we can do the same with the point below 3. The slope there (as calculated above, is 5.2. So we can approximate the instantaneous velocity at t = 3 seconds by averaging the two: (6.9 + 5.2)/2 = 6.05 m/s.

To curve fit the data, plug it into a list in your TI-84 calculator and try the various types of curves to find the one with the best r² value. A quartic function gives the r² closest to 1 (r² = 0.99993). It is:

s(t) = 0.00625t⁴ - 0.1300925926t³ + 1.475694444t² + 0.1171957672t - 0.0115079365

Using this function and using the limit equation, for the following values of "h" I get values closer and closer to the instantaneous velocity:

For h = 0.1, v = 6.19755 m/s

For h = 0.01, v = 6.14028 m/s

For h = 0.001, v = 6.1345 m/s

For h = 0.0001, v = 6.1339 m/s

For h = 0.00001, v = 6.1339 m/s

For h = 0.000001, v = 6.1339 m/s

So the approximation of the instantaneous velocity, at t = 3 seconds, using this curve-fit method is 6.1339 m/s