A particle at rest leaves the origin with its velocity increasing with time according to v₁(t) = 3.3t m/s. At 5 s, the particles velocity starts decreasing according to v₂(t) = 16.5 1.8(t-5) m/s. The decrease continues until t = 11s, after which the particle's velocity remains constant at v3(t) = 5.7 m/s. Help on how to format answers: units (a) What is the acceleration of the particle as a function of time? 0≤ t ≤ 5: 5 < t < 11: 11

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A particle at rest leaves the origin with its velocity increasing with time according to v₁(t) = 3.3t m/s. At 5 s, the particles velocity
starts decreasing according to v₂(t) = 16.5 - 1.8(t - 5) m/s. The decrease continues until t = 11s, after which the particle's velocity
remains constant at v3(t) = 5.7 m/s.
Help on how to format answers: units
(a) What is the acceleration of the particle as a function of time?
0≤t≤ 5:
5 < t < 11:
11<t:
(b) What is the position of the particle at:
t = 4s:
t=7s:
t = 13 s:
Transcribed Image Text:A particle at rest leaves the origin with its velocity increasing with time according to v₁(t) = 3.3t m/s. At 5 s, the particles velocity starts decreasing according to v₂(t) = 16.5 - 1.8(t - 5) m/s. The decrease continues until t = 11s, after which the particle's velocity remains constant at v3(t) = 5.7 m/s. Help on how to format answers: units (a) What is the acceleration of the particle as a function of time? 0≤t≤ 5: 5 < t < 11: 11<t: (b) What is the position of the particle at: t = 4s: t=7s: t = 13 s:
Expert Solution
Introduction

The initial velocity of the particle is v1t=3.3t m/s.

After 5 s the velocity of the particle becomes v2t=16.5-1.8t-5 m/s.

After 11 s the velocity of the particle becomes v3t=5.7 m/s.

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