Question

Below is a set of data that represents the velocity (in feet per second) of the final 45 seconds of the landing. At t 0, the plane is on its final descent. Table I t in seconds 10 20 25 30 45 5 15 35 40 v(t) feet per second 274.27 223.19 179.23 108.83 80.80 56.68 35.91 18.04 2.65 141.4 Table II t in seconds 4 5 14 15 24 25 34 35 44 45 v(t) in feet per second 148.52 86.08 80.80 39.82 35.91 5.55 2.65 232,8 223.19 141.4 Part ll: Analysis of Data Applying Derivatives Calculating average acceleration. A. Using the data in Table I, calculate the average acceleration for the following intervals: From t 0 to t 45 i. From t 25 to t 45 ii From t 40 to t 45 Calculating instantaneous acceleration. B. 1. Using the data in Table II, calculate the instantaneous acceleration at the following intervals: i. t= 5 ii. t= 15 iii. t 25 iv t = 35 v. t 45 Explain how you used the limit definition of a derivative to calculate the instantaneous acceleration. Use your results to explain why the limit definition of a derivative is true. 2. At what point is the acceleration at a maximum? How is this relevant to the landing aircraft? 3. For each of the questions above, provide supporting solutions and calculus terminology to explain how you arrived at your answers. Also, explain in detail what your answer represents in a real-world context and why this is useful.

Answer 1

A. (i) Fromt =0 to t = 45 Average acceleration - V(45)-v(0) 45-0 2.65 274.27 =-6.036 ft/s4 - 45-0 (ii From t25 to t 45 v(45)v(25) 2.65 80.80 =-3.9075 ft/s Average acceleration = 20 45 25 (iiiFromt=40 to t = 45 v(45)-v(40) 2.65 18.04 =-3.078 ft/s Average acceleration = 45-40 5 B. (1.) Here the instantaneous acceleration is given by, a(t) )(t-1) = 1 (i Att 5 a(5)= "(5)-v(4). = 223.19 232.8 -9.61 ft/s 1 v(15)-v(14) (ii) Att 15, a(15): - 141.4-148.52 =-7.12 ft/s- 1 = 80.80 -86.08 =-5.28 ft/s (iii Att 25, a(25)- "(25)-v(24) 1 v(35 (34) = (iv) Att 35, a(35) - 35.91-39.82 -3.91 ft/s- 1 v(45)- г(44) (v) Att = 45, a(45) = 2.65-5.55 -2.9 ft/s = 1 The derivative of a function f (x) is defined as (2.) f'(x)= limJ (.x)-f(x-h) where h-»0 means h is very small h0 It gives the instantaneous rate of change, therefore, we have used it in our problem to find instantaneous acceleration, where f(x) as v(t) and h 1 as very small quantity. (3.) Considering the negative sign as indicating the downward direction, the acceleration is maximum at t = 5. It is very relevant to landing aircraft, as with increasing t, the magnitude of acceleration is decreasing. Please rate this answer. Thanks