A particle starts from rest at a fixed point 0 and its acceleration is 12t^2-24t+8 m/s^2. Show that the particle is again at rest when it returns to 0 and find its maxi- mum displacement from 0 before it returns. Inte

Accepted Solution

Answer:a) V(2) = 0 The particle is at restb) The maximum displacement is 1mStep-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=0mNow, 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 = 2Since x(0) and x(2) are both 0, we calculate x(1) to find the maximum displacement:x(1) = 1m