We are given the following data for the velocity as a function of time: 10 227.04 362.78 517.35 602.97 | 901.67 15 20 22.5 t(s) v(t) (m/s) 30 (1) Determine the value of the velocity at t-16 s using third order Lagrangian polynomial interpolation. (2) Find the absolute relative approximate error for the third order polynomial approximation, if v(16) = 392.19 m/s using the second order polynomial. (3)Using the third order polynomial interpolant for velocity, find the distance covered from t-11 s tot=16 s. (4) Using the third order polynomial interpolant for velocity, find the acceleration at t=16 s.

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 91E
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We are given the following data for the velocity as a function of time:
20
227.04 362.78 517.35 602.97 901.67
10
30
t(s)
v(t) (m/s)
15
22.5
(1) Determine the value of the velocity at t=16 s using third order Lagrangian polynomial interpolation.
(2) Find the absolute relative approximate error for the third order polynomial approximation, if v(16) =
392.19 m/s using the second order polynomial.
(3)Using the third order polynomial interpolant for velocity, find the distance covered from t-11 s to t=16 s.
(4) Using the third order polynomial interpolant for velocity, find the acceleration at t=16 s.
Transcribed Image Text:We are given the following data for the velocity as a function of time: 20 227.04 362.78 517.35 602.97 901.67 10 30 t(s) v(t) (m/s) 15 22.5 (1) Determine the value of the velocity at t=16 s using third order Lagrangian polynomial interpolation. (2) Find the absolute relative approximate error for the third order polynomial approximation, if v(16) = 392.19 m/s using the second order polynomial. (3)Using the third order polynomial interpolant for velocity, find the distance covered from t-11 s to t=16 s. (4) Using the third order polynomial interpolant for velocity, find the acceleration at t=16 s.
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