The Distance the Particle Travels with Given Position Function x(t) = t^4 - 8t^2 | CE Board Problem in Mathematics, Surveying and Transportation Engineering

Date of Exam

May 2016

Subject

Problem
Given the position function x(t) = t⁴ - 8t², find the distance that the particle travels at t = 0 to t = 4.

A. 160	C. 140
B. 150	D. 130

Answer Key

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[ A ]

Solution

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$\displaystyle s = \int_{t_1}^{t_2} \sqrt{1 + \left( \dfrac{dx}{dt} \right)^2} \, dt$

$\displaystyle s = \int_0^4 \sqrt{1 + (4t^3 - 16t)^2} \, dt$

$s = 160.9$ ← Answer: [ A ]

Solution by Simpson's One-third Rule:

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$f(t) = \sqrt{1 + (4t^3 - 16t)^2}$

Let
$\Delta = \dfrac{4 - 0}{10} = 0.4$

n	t	f(t)	C	C * f(t)
1	0	1.00	1	1.00
2	0.4	6.22	4	24.90
3	0.8	10.80	2	21.60
4	1.2	12.33	4	49.31
5	1.6	9.27	2	18.54
6	2	1.00	4	4.00
7	2.4	16.92	2	33.85
8	2.8	43.02	4	172.08
9	3.2	79.88	2	159.76
10	3.6	129.03	4	516.11
11	4	192.00	1	192.00
Sum =				1193.15

Simpson's One-third Rule may be written as
$s = \dfrac{\Delta}{3} \Big[ f(t)_1 + 2\Sigma f(t)_{odd} + 4 \Sigma f(t)_{even} + f(t)_n \Big]$

From the table above
$f(t)_1 + 2\Sigma f(t)_{odd} + 4 \Sigma f(t)_{even} + f(t)_n$ = Sum

Hence,
$s = \dfrac{0.4}{3}(1193.15) = 159.09$ ← answer