A spreadsheet is an excellent way to do such an assignment. Here is an Excel spreadsheet with the 8 parts of the assignment done. As you can see, point P remains the same throughout the assignment. Point Q varies according to the assignment. The y-coordinates for point Q are calculated using the built in log function (sample: +ln(E2)). The slope is calculated using a formula like: (+(F12-C12)/(E12-B12)). 

Just looking at the results from part (a) we see that the closest point to 2 is part (iv). The answer for this portion of part (a) was very close to .5 

Taking the slope-intercept form of a line, we substitute in the point (2,ln2) and solve for b. 
 

Therefore, this is our estimate of the equation of the tangent line at that point. 

Using Maple, we can check if this makes sense. We use the following Maple command: 

> plot([ln(x),.5*x-.3069],x=0..4);

This gives us both the graph of the function and the tangent line at the desired point. 

We see that the result is consistent with our estimate. The tangent line appears to be tangent to the funciton at the point (2,ln2). The slope appears to be about +.5 and the y-intercept appears to be about -.3. 

Here we see the sketch of the curve (red), the tangent line (green) and two of the secant lines (blue and yellow). 

We were to start at t = 2. So our first point is (2,32). The time period is 3 seconds. So our second point is (5,178). Using these two points, calculate the average rate of change or slope.

First, plot the points in the table. Then draw in your estimate of the tangent line at t = 2. Calculate an approximate slope for the line.

(I used (3,60) and (1,5) as two points on the line to calculate the slope from).

I got about 27.5 or 28 feet/second.