NOCHILLL
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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

Respuesta :

the particle's initial position is at t=0, x = 0 - 0 + 4 = 4m

velocity is rate of change of displacement = dx/dt = d(t^3 - 9t^2 +4)/dt
= 3t^2 - 18t

acceleration is rate of change of velocity = d(3t^2 -18t)/dt
= 6t - 18

the particle is stationary when velocity = 0, so 3t^2 - 18t =0
3t*(t - 6) = 0
t = 0 or t = 6s

acceleration = 6t - 18 = 0
t = 3s
at t = 3s, velocity = 3(3^2) -18*3 = -27m/s
displacement = 3^3 - 9*3^2 +4 = -50m

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

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