Will give brainliest the displacement, x meters, from the origin at time, t seconds, of a particle moving in a straight line is given by x=t^3 -9t^2 +4 (i) find the particle's initial position. (ii) find an expression for the velocity as a function of time. (iii) find an expression for the acceleration as a function of time. (iv) at what times is the particle stationary? (v)at what time is the acceleration zero? find its velocity and position then

i) t=0, x = 4

ii) velocity = dx/dt = 3t^2 - 18t

iii) acceleration = d(velocity)/dt = 6t - 18

iv) vel=0 when particle is stationary, 3t^2 - 18t =0

t = 0 or 6

v) 6t - 18 = 0

t = 3

vel = 3(3^2) -18*3 = -27

position = 3^3 - 9*3^2 +4 = -50

u = -(6/7)w

step-by-step explanation:

place all values with 'u' on one side of the equation.

4u + 8w = - 3u + 2w

4u + 3u = 2w - 8w

further simplify the equation by solving the addition and subtraction.

7u = - 6w

make 'u' the independent variable by placing its co-efficient on the right-hand side.

u = - (6/7)w