TABLE 16.3 Kelly Schmitzer skating data t (sec) s (m) t (sec) s (m) t (sec) s (m) 1.5 0.89 3.1 1.30 0.1 0.07 1.7 0.97 3.3 1.31 0.3 0.22 1.9 1.05 3.5 1.32 0.5 0.36 2.1 1.11 3.7 1.32 0.7 0.49 2.3 1.17 3.9 1.32 0.9 0.60 2.5 1.22 4.1 1.32 1.1 0.71 2.7 1.25 4.3 1.32 1.3 0.81 2.9 1.28 4.5 1.32

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question

Coasting to a stop Table 16.3 shows the distance s (meters)
coasted on in-line skates in terms of time t (seconds) by Kelly
Schmitzer. Find a model for her position in the form.

Her initial velocity was v0= 0.80 m/sec,  her mass m = 49.90 kg (110 lb), and her total coasting distance was 1.32 m.

TABLE 16.3 Kelly Schmitzer skating data
t (sec)
s (m)
t (sec)
s (m)
t (sec)
s (m)
1.5
0.89
3.1
1.30
0.1
0.07
1.7
0.97
3.3
1.31
0.3
0.22
1.9
1.05
3.5
1.32
0.5
0.36
2.1
1.11
3.7
1.32
0.7
0.49
2.3
1.17
3.9
1.32
0.9
0.60
2.5
1.22
4.1
1.32
1.1
0.71
2.7
1.25
4.3
1.32
1.3
0.81
2.9
1.28
4.5
1.32
Transcribed Image Text:TABLE 16.3 Kelly Schmitzer skating data t (sec) s (m) t (sec) s (m) t (sec) s (m) 1.5 0.89 3.1 1.30 0.1 0.07 1.7 0.97 3.3 1.31 0.3 0.22 1.9 1.05 3.5 1.32 0.5 0.36 2.1 1.11 3.7 1.32 0.7 0.49 2.3 1.17 3.9 1.32 0.9 0.60 2.5 1.22 4.1 1.32 1.1 0.71 2.7 1.25 4.3 1.32 1.3 0.81 2.9 1.28 4.5 1.32
Expert Solution
Step 1: Introduction

As per the question, we are tasked with finding a mathematical model for Kelly Schmitzer's position in terms of time while coasting on in-line skates. We are provided with a table showing the distance she coasted in meters for various time intervals.

Using this data, along with her initial velocity, mass, and total coasting distance, we will determine a quadratic equation that represents her position as a function of time.

By fitting the data points to the equation, we can find the coefficients and establish a model that describes Kelly's motion while coasting.

Step 2: Calculation

Advanced Math homework question answer, step 2, image 1

Python Code :

import numpy as np

 

# Given data points

t = np.array([1.5, 1.7, 1.9, 2.1, 2.3, 2.5, 2.7, 2.9])

s = np.array([0.89, 0.97, 1.05, 1.11, 1.17, 1.22, 1.25, 1.28])

 

# Construct the matrix A and vector b

A = np.vstack([t**2, t]).T

b = s

 

# Use linear regression to solve the system of equations

coefficients, residuals, _, _ = np.linalg.lstsq(A, b, rcond=None)

 

# Extract the coefficients

a = coefficients[0]

b = coefficients[1]

 

# Display the coefficients

print(f"a: {a}")

print(f"b: {b}")

 

Code Output :

a: -0.10881418726475356

b: 0.7577663226819888

[Program finished]

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