# Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x'(t) is its velocity, and x"(t) is its acceleration

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x'(t) is its velocity, and x"(t) is its acceleration.
$x\left(t\right)={t}^{3}-6{t}^{2}+9t-9,$
$0\le t\le 10$
a) Find the velocity and acceleration of the particle?
${x}^{\prime }\left(t\right)=?$
$x"\left(t\right)=?$
b) Find the open t-intervals on which the particle is moving to the right?
c) Find the velocity of the particle when the acceleration is 0?
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Clelioo

Step 1
Position as a function of time is given. We have to find the velocity and acceleration.
Recall the famous rule of differentiation:
$d\left({x}^{n}\right)/dx=n{x}^{n-1}$
Part (a)
Velocity $={x}^{\prime }\left(t\right)=3{t}^{2}-12t+9$
Acceleration $=x"\left(t\right)+6t-12$
Part (b)
When particle is moving to the right, velocity $>0$
Hence, ${x}^{\prime }\left(t\right)>0$
Hence, $3{t}^{2}-12t+9>0$
Or, ${t}^{2}-4t+3>0$
Or, $\left(t-3\right)\left(t-1\right)>0$
Hence,
The bounds for t as stated in question is: $0\le t\le 10$
Hence, the open t-interval will be: $\left[0,1\right)\cup \left(3,10\right]$
Part (c)
Acceleration is zero when $x"\left(t\right)=6t-12=0$
Hence, $t=12/6=2$
Hence, velocity when acceleration is zero $={x}^{\prime }\left(t=2\right)=3\cdot {2}^{2}-12\cdot 2+9=-3$