a) Solve your differential equation for v(t) b) If you choose to solve the equation in v(t), the velocity, then you can integrate the velocity function you obtained, v(t), above to get position function
a) Solve your differential equation for v(t) b) If you choose to solve the equation in v(t), the velocity, then you can integrate the velocity function you obtained, v(t), above to get position function
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
**coffee filter has mass 0.01908 kg
total stack mass m = 12 × 0.01908 = 0.22896 kg
AveVelInitial = y3 − y1/t3-t1 = 1.10446 m/s
t = 0 at the second observation of our collected data
distance fallen here is 0 m
initial velocity of 1.10446 m/s.
v(t) = (g−1/2m*ρACvo2)t+c
Where variables are
g is acceleration due to gravity
ρ is density of airC is drag coefficient
A is the cross sectional area of the object
vo is the initial velocit
m is mass of stacked coffee filters **
a) Solve your differential equation for v(t)
b) If you choose to solve the equation in v(t), the velocity, then you can integrate the velocity function you obtained, v(t), above to get position function
Transcribed Image Text:Time (s)
Distance (m)
Time (s)
Distance (m)
0.00
0.0000
0.22
0.3619
0.02
0.0228
0.24
0.4050
0.04
0.0478
0.26
0.4500
0.06
0.0730
0.28
0.4931
0.08
0.1054
0.30
0.5440
0.10
0.1369
0.32
0.5927
0.12
0.1710
0.34
0.6429
0.14
0.2063
0.36
0.6937
0.16
0.2426
0.38
0.7458
0.18
0.2810
0.40
0.7991
0.20
0.3197
0.42
0.8556
Table 1. Data on the time and position of a stack of coffee filters as it falls to the
ground. We measure the distance falling positively downward and say y(t) is the
distance fallen in meters at time t in seconds, i.e. t is time in seconds from release
and y(t). is distance in meters fallen from release.
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