Initially, a bicycle is at the position ř = (130 m)î + (210 m)j. After 120 s, the position of the particle is ř = (110 m)î + (240 m)j. What is the average velocity vector of the bicycle in that time tinterval? O Tay = (0.17 m/s))î + (–0.25 m/s)ĵ O ūay = (0.25 m/s)î + (–0.17 m/s)j O Tay = (-0.25 m/s)î + (0.17 m/s)j O Tay = (-0.17 m/s)î + (0.25 m/s)j

Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter3: Motion In Two Dimensions
Section: Chapter Questions
Problem 3P: A particle initially located at the origin has an acceleration of a=3.00jm/s2 and an initial...
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Initially, a bicycle is at the position 7 = (130 m)i + (210 m)j. After 120 s, the position of the particle is 7 = (110 m)i + (240 m)j . What is the average velocity vector of the bicycle in that time
||
tinterval?
O Tav = (0.17 m/s)i + (-0.25 m/s)j
O Jay = (0.25 m/s)i + (-0.17 m/s)
O Tay = (-0.25 m/s)î + (0.17 m/s)
O Jay = (-0.17 m/s)i + (0.25 m/s)j
Transcribed Image Text:Initially, a bicycle is at the position 7 = (130 m)i + (210 m)j. After 120 s, the position of the particle is 7 = (110 m)i + (240 m)j . What is the average velocity vector of the bicycle in that time || tinterval? O Tav = (0.17 m/s)i + (-0.25 m/s)j O Jay = (0.25 m/s)i + (-0.17 m/s) O Tay = (-0.25 m/s)î + (0.17 m/s) O Jay = (-0.17 m/s)i + (0.25 m/s)j
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