The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s = 2 sin(πt) + 5 cos(πt), where t is measured in seconds. A) Find the average velocity during each time period.
1) [1, 2]
2) [1, 1.1]
3) [1, 1.01]
4) [1, 1.001]
B) Estimate the instantaneous velocity of the particle when t = 1. cm/s

  A) 10, -3.73, -6.035, -6.259 . . . cm/s

  B) -6.2832 cm/s

Step-by-step explanation:

A) For problems like this, where repeated evaluation of a function is required, I find a graphing calculator or spreadsheet to be an appropriate tool. The attached shows that we defined the position function ...

  p(t) = 2sin(πt) +5cos(πt)

and a function for computing the average velocity from t=1. For some time interval ending at t2, the average velocity is ...

  Va(t2) = Δp/Δt = (p(t2) -p(1))/(t2 -1)

Then, for example, for t2 = 2, the average velocity on the interval [1, 2] is ...

  Va(2) = (p(2) -p(1))/(2 -1) = ((2sin(2π) +5cos(2π)) -(2sin(π) +5cos(π)))/(1)

  = (2·0+5·1 -(2·0 +5·(-1)) = 10 . . . . matches the table value for x1 = 2.

Then the average velocity values for the intervals of interest are ...

  1) [1, 2]   Va = 10

  2) [1, 1.1]   Va = -3.73

  3) [1, 1.01]   Va = -6.035

  4) [1, 1.001]   Va = -6.259

__

B) Sometimes a better estimate is obtained when the interval is centered on the point of interest. Here, we can compute the average velocity on the interval [0.999, 1.001] as a better approximation of the instantaneous velocity at t=1. That value is ...

  [0.999, 1.001]   Va = -6.283175*

Our estimate of V(1) is -6.2832 cm/s.

The exact value is -2π ≈ -6.2831853... cm/s

__

* This is the average of the Va(0.999) and Va(1.001) values in the table.

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents