Answer:
a) V(2) = 0 The particle is at rest
b) The maximum displacement is 1m
Step-by-step explanation:
Let's first find velocity and displacement from the given acceleration:
[tex]v(t) = \int\limits {a(t)} \, dt = 4t^3-12t^2+8t[/tex]
[tex]x(t) = \int\limits {v(t)} \, dt = t^4-4t^3+4t^2[/tex]
Now we need the instant t1 where x(t1) = 0 and then we evaluate v(t1) to find its velcity and verify if it is 0:
[tex]x(t1) = 0 = t1^4-4t1^3+4t1^2 = t1^2 * (t1^2 - 4t1 + 1)[/tex] Solving for t, we get:
t1 = 0 and t1 = 2s Now we evaluate v(2):
v(2) = 0m/s It is at rest when it returns to x=0m
Now, for maximum displacement, we need the instant when v(tm) = 0
[tex]v(tm) = 4tm^3-12tm^2+8tm = t * (4*tm^2 - 12*tm + 8) = 0[/tex] Solving for tm:
tm=0; tm = 1 and tm = 2
Since x(0) and x(2) are both 0, we calculate x(1) to find the maximum displacement:
x(1) = 1m
