Calculating instantaneous acceleration.

Question

1. Using the data, calculate the instantaneous acceleration at the following intervals:

i.             t = 5

ii.            t = 15

iii.           t = 25

iv.           t = 35

v.            t = 45

table of data

t in seconds                     

4              5              14             15                24            25            34            35            44            45

v(t) in feet per second       

232.8       223.19     148.52        141.4       86.08       80.80       39.82       35.91        5.55         2.65

2.at what point is the acceleration at a maximum?

Answer 1

I would be tempted to do a least squares curve fit and use the function to find the derivative at each of the points in question but I doubt if that's something you would be expected to do.

To find the slope at t = 5, I would find the slope from t = 4 to 5, then from 5 to 14, and then average them.

The slope from 4 to 5 is (232.8 - 223.19)/(4 - 5) = -9.61

The slope from 5 to 14 is (223.19 - 148.52)/(5 - 14) = -8.30

The average is -8.95 ft/s²

Your teacher may just expect you to give the slope at t=5 from the -9.61 value because the point at t=5 is very close to the two points t = 4 and t = 5.

Using the curve-fit method I mentioned earlier gives the slope at -8.74 ft/s² which is fairly close to the average.

Whichever way you decide to do it, you just repeat the process for the other data points.