Answer:
a) Right 0≤p<π/2 and 3π/2<p≤2π
left π/2 ≤ p ≤3π/2
stopped p=π/2 and p=3π/2
b) 5.5cm and its final position (2π, 0)
c) 6.5cm
Step-by-step explanation:
we must find where the function becomes negative, positive and zero
(according to the graph)
the particle moves to the right where the function is positive
0≤p<π/2 and 3π/2<p≤2π , the particle moves to the left where the function is negative π/2 ≤ p ≤3π/2, and stopped p=π/2 and p=3π/2
s(t)=∫v(t)dt also s(t) = 5sin(t)
(according to the graph 2)
The displacement of the particle is 5.5 and its final position
(2π, 0)
The total displacement of the particle is 6.5
