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4. Use a third order polynomial approximation of s(t) to approximate the velocity at t=30s using a. Direct Fit Polynomial Method b. Lagrange Polynomial Method

4. Use a third order polynomial approximation of s(t) to approximate the velocity at t=30s using a. Direct Fit Polynomial Method b. Lagrange Polynomial Method

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
Chapter2: Graphical And Tabular Analysis
Section2.1: Tables And Trends
Problem 1TU: If a coffee filter is dropped, its velocity after t seconds is given by v(t)=4(10.0003t) feet per...
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pls help me with number 4 thank you 

The distance s, measured in meters, of a bicycle rider from his point of origin is measured at intervals of 0 25 seconds, as shown below 3. 0.25 0.5 0.75 1.25 1.5 4.3 10.2 17.2 26.2 33.1 39.1 Use CDD to approximate the velocity and acceleration of the biker at t=0.25 s, 0.75s and 1.25 s. Give your answers in 4 decimal places. Time Velocity Acceleration 0.25 s 0.75 s 1.25 s s (0. 75) = ao + 0. 75a, + a2 + az = 17.2 = ao + 0.75a + az + az 1642 s (1) = ao + aj + az + az > 26. 2 = ao + aị + a2 + az 25 s (1. 25) = ao + 1. 25ai + a2 + = 33. 1 = ao + 1. 25a, + a 1642 + 125 64 a3 s (1.5) = ao + 1. 5a1 + a2 + az = 39.1 = ao + 1. 5a1 + a2 + az. This system of equations can be written in the matrix form, 1 0.75 17.2 1 1 1 1 26. 2 1 1. 25 25 125 33. 1 a2 64 1 1.5 39. 1 We reduce this matrix using Gaussian Elimination, 1 0.75 /16 27/64 17.2 1 1 1 1 26. 2 1 1.25 25/16 125/64 33. 1 1 1.5 4 27/s 39. 1 R2 → R2 – R1, R3 → R3 – R1, R4 → R4 – R1 1 0. 75 0. 563 0.422 17. 2 0 0. 25 0.438 0. 578 9 1. 531 15.9 0 0. 75 1.688 2.953 21.9 0 0.5
Transcribed Image Text:The distance s, measured in meters, of a bicycle rider from his point of origin is measured at intervals of 0 25 seconds, as shown below 3. 0.25 0.5 0.75 1.25 1.5 4.3 10.2 17.2 26.2 33.1 39.1 Use CDD to approximate the velocity and acceleration of the biker at t=0.25 s, 0.75s and 1.25 s. Give your answers in 4 decimal places. Time Velocity Acceleration 0.25 s 0.75 s 1.25 s s (0. 75) = ao + 0. 75a, + a2 + az = 17.2 = ao + 0.75a + az + az 1642 s (1) = ao + aj + az + az > 26. 2 = ao + aị + a2 + az 25 s (1. 25) = ao + 1. 25ai + a2 + = 33. 1 = ao + 1. 25a, + a 1642 + 125 64 a3 s (1.5) = ao + 1. 5a1 + a2 + az = 39.1 = ao + 1. 5a1 + a2 + az. This system of equations can be written in the matrix form, 1 0.75 17.2 1 1 1 1 26. 2 1 1. 25 25 125 33. 1 a2 64 1 1.5 39. 1 We reduce this matrix using Gaussian Elimination, 1 0.75 /16 27/64 17.2 1 1 1 1 26. 2 1 1.25 25/16 125/64 33. 1 1 1.5 4 27/s 39. 1 R2 → R2 – R1, R3 → R3 – R1, R4 → R4 – R1 1 0. 75 0. 563 0.422 17. 2 0 0. 25 0.438 0. 578 9 1. 531 15.9 0 0. 75 1.688 2.953 21.9 0 0.5
R3 → R3 – 2R2, R4 → R4 – 3R2 1 0.75 0. 563 0.422 17.2 0 0. 25 0. 438 0. 578 9 0. 125 0. 375 -2.1 0. 375 1.219 -5.1 R4 - R4 - 3R3 1 0.75 0. 563 0, 422 | 17.2 0 0. 25 0.438 0. 578 0. 125 0.375|-2.1 0.094 1.2 From above, we can calculate the coefficients, we get do = -34. 4 aj = 103 az = -55.2 az = 12. 8. The required polynomial will be s (1) = -34. 4 + 1031 – 55. 21² + 12. 8,3. Differentiating the obtained polynomial, we get v (1) = s' (1) = 103 – 110. 41 + 38. 412 + v (30) = 103 – 110. 4 (30) + 38. 4(30)? > v (30) = 31351m/s. → v (0. 3) = 103 – 110.4 (0. 3) + 38. 4(0. 3)² → v (0. 3) = 73m/s. 4. Use a third order polynomial approximation of s(t) to approximate the velocity at t=30s using Direct Fit Polynomial Method b. Lagrange Polynomial Method a.
Transcribed Image Text:R3 → R3 – 2R2, R4 → R4 – 3R2 1 0.75 0. 563 0.422 17.2 0 0. 25 0. 438 0. 578 9 0. 125 0. 375 -2.1 0. 375 1.219 -5.1 R4 - R4 - 3R3 1 0.75 0. 563 0, 422 | 17.2 0 0. 25 0.438 0. 578 0. 125 0.375|-2.1 0.094 1.2 From above, we can calculate the coefficients, we get do = -34. 4 aj = 103 az = -55.2 az = 12. 8. The required polynomial will be s (1) = -34. 4 + 1031 – 55. 21² + 12. 8,3. Differentiating the obtained polynomial, we get v (1) = s' (1) = 103 – 110. 41 + 38. 412 + v (30) = 103 – 110. 4 (30) + 38. 4(30)? > v (30) = 31351m/s. → v (0. 3) = 103 – 110.4 (0. 3) + 38. 4(0. 3)² → v (0. 3) = 73m/s. 4. Use a third order polynomial approximation of s(t) to approximate the velocity at t=30s using Direct Fit Polynomial Method b. Lagrange Polynomial Method a.
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