In class, we used a numerical method to determine the velocity of an object at t = 2.00 s given that the object had started from rest at t= 0s and accelerated at a rate given by a(t) = (8.00 m/s) ť². Here's what we did: 1. Divided t into four intervals of 0.5 s each. 2. For each interval, calculated the average a. These were the numerical values we obtained for each time interval: i t (s) a, (m/s²) After summing the values according to: 1 0.25 0.50 2 0.75 4.50 3 1.25 12.50 V₂ -v₁ = lima At 4 we arrived at the approximate value v = 21.0 m/s. 4 1.75 24.50 a (m/s²) 32.00+ 24.00+ 16.00 8.00+ ā az az a4 0 0 0.50 1.00 1.50 2.00 -t (s) We then proceeded to integrate to obtain the analytical solution, which was v = 21.33 m/s. Using the same numerical method, divide t into eight and then sixteen intervals to calculate the approximate value for v at t = 2.00 s. This should demonstrate that with more sampling, the AUC (area under the curve) value converges toward the analytical solution. Make two tables similar to the one above, showing the t and a values for each interval, and then sum the areas.

Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter1: Getting Started
Section: Chapter Questions
Problem 18PQ: Acceleration a has the dimensions of length per time squared, speed v has the dimensions of length...
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In class, we used a numerical method to determine the velocity of an object at
0s and
=
t = 2.00 s given that the object had started from rest at t
accelerated at a rate given by a(t) = (8.00 m/s^)t². Here's what we did:
1. Divided t into four intervals of 0.5 s each.
2. For each interval, calculated the average a.
These were the numerical values we obtained for each time interval:
1
0.25
0.50
i
t (s)
a; (m/s²)
After summing the values according to:
2
0.75
4.50
3
1.25
12.50
lim ã‚At,
2
V₂v₁ = limast.
t₁
At→0
we arrived at the approximate value v = 21.0 m/s.
4
1.75
24.50
a (m/s²)
32.00+
24.00
16.00
8.00
0
ā₁
0
ā2
az
as
0.50 1.00 1.50 2.00
t (s)
21.33 m/s.
We then proceeded to integrate to obtain the analytical solution, which was v =
Using the same numerical method, divide t into eight and then sixteen intervals to calculate the approximate
value for v at t = 2.00 s. This should demonstrate that with more sampling, the AUC (area under the curve)
value converges toward the analytical solution. Make two tables similar to the one above, showing the t and a
values for each interval, and then sum the areas.
Transcribed Image Text:In class, we used a numerical method to determine the velocity of an object at 0s and = t = 2.00 s given that the object had started from rest at t accelerated at a rate given by a(t) = (8.00 m/s^)t². Here's what we did: 1. Divided t into four intervals of 0.5 s each. 2. For each interval, calculated the average a. These were the numerical values we obtained for each time interval: 1 0.25 0.50 i t (s) a; (m/s²) After summing the values according to: 2 0.75 4.50 3 1.25 12.50 lim ã‚At, 2 V₂v₁ = limast. t₁ At→0 we arrived at the approximate value v = 21.0 m/s. 4 1.75 24.50 a (m/s²) 32.00+ 24.00 16.00 8.00 0 ā₁ 0 ā2 az as 0.50 1.00 1.50 2.00 t (s) 21.33 m/s. We then proceeded to integrate to obtain the analytical solution, which was v = Using the same numerical method, divide t into eight and then sixteen intervals to calculate the approximate value for v at t = 2.00 s. This should demonstrate that with more sampling, the AUC (area under the curve) value converges toward the analytical solution. Make two tables similar to the one above, showing the t and a values for each interval, and then sum the areas.
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