1. A particle moving along the x-axis in such a way that its position at time t is given by x=3t^4 -16t^3+24t^?

a/ v= 12t^3 - 48t^2 + 48t = 12t(t-2)^2 a = 36t^2 - 96t + 48 b/ v = 0 ---> t = 0 , 2 c/ when t = 0 d/ a = 0 ---> t = 2/3 and v(2/3) = 128/9

Here we go, x=3t^4 -16t^3+24t^2 a. Differentiate both sides by t, Velocity, dx/dt = 12 t^3 - 48 t^2 + 48 t --------- (1) Acceleration, d^2x/dt^2 = 36 t^2 - 96 t + 48 ----------- (2) b. here V = 0; from 1st eqn, 0 = 12 t^3 - 48 t^2 + 48 t so t = 0 or 2 s c. at t = 0 (here it should be either at 0 or 2 or both, but by applying some values we can see it should be 0) d. here consider eqn 2 = 0; that gives t = 2/3 or 2. Since 2/3 < 2, consider the occation when t = 2/3, then apply that value to eqn 1, to get the answer, V = 128/9 (i.e. 14.22 m/s).

x=3t^4 -16t^3+24t^2 [a] v = 12t^3 - 48t^2 + 48t = 12t(t - 2)^2 a = 36t^2 - 96t + 48 [b] v = 12t^3 - 48t^2 + 48t = 0 Particle is at rest at t = 0, and at t = 2. [c] For a =0, t = ? a = 36t^2 - 96t + 48 = 0 t = 2/3 sec and at t = 2 sec. for furst a = 0, t = 2/3 sec

x(t) = 3t^4 - 16t³ + 24t² a) v(t) = dx/dt = 12t³ - 48t² + 48t a(t) = dv/dt = 36t² - 96t + 48 b) at rest when v(t) = 0 12t³ - 48t² + 48t = 0 12t(t² - 4t + 4) = 0 12t(t - 2)² = 0 at rest when t = 0 and when t = 2 a(0) = 48 > 0 changes direction at t = 0 a(2) = 36t(2)² - 96(2) + 48 = 0 (this is a point of inflection) only at t = 0 d) a(t) = 0 36t² - 96t + 48 = 0 12(3t² - 8t + 4) = 0 12(3t - 2)(t - 2) = 0 a(t) = 0 when t = 2 and t = 2/3 v(2/3) = 12(2/3)³ - 48(2/3)² + 48(2/3) = 14.2222....