The velocity function is v(t) = t² — 5t + 4 for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time interval [-2,5]. displacement = (5² (55²) +4-5)-((-2³ (5-(-2²) +4- 3 2 distance traveled = 35 +4·(-2)
The velocity function is v(t) = t² — 5t + 4 for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time interval [-2,5]. displacement = (5² (55²) +4-5)-((-2³ (5-(-2²) +4- 3 2 distance traveled = 35 +4·(-2)
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
Chapter2: Graphical And Tabular Analysis
Section2.1: Tables And Trends
Problem 1TU: If a coffee filter is dropped, its velocity after t seconds is given by v(t)=4(10.0003t) feet per...
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![The velocity function is v(t) = t² – 5t + 4 for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time
interval [-2,5].
displacement = ( (532) +-4-5)-(4-2² (3-(-2²) +4·(-2))
(5-(−2)²)
distance traveled = 35](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F89e7fc9b-2d74-4b86-bca1-07d9b67ff60f%2F53d8b333-b959-4923-bc20-b2890557389f%2Fix2bipp_processed.png&w=3840&q=75)
Transcribed Image Text:The velocity function is v(t) = t² – 5t + 4 for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time
interval [-2,5].
displacement = ( (532) +-4-5)-(4-2² (3-(-2²) +4·(-2))
(5-(−2)²)
distance traveled = 35
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