Question

The position of a particle moving along the x-axis is given by `x(t)=t^3-9t^2+24t` meters where t is in seconds, `t>=0` a) Draw a sign diagram for the particle's velocity and acceleration functions.?

`t>=0`

Answer 1

See the explanation below

Explanation:

The velocity is the derivative of the position

`x(t)=t^3-9t^2+24t`

`v(t)=3t^2-18t+24=3(t^2-6t+8)`

`=3(t-2)(t-4)`

The acceleration is the derivative of the velocity

`v(t)=3t^2-18t+24`

`a(t)=6t-18=6(t-3)`

The sign chart for the velocity is as follows :

`color(white)(aaaa)` `t` `color(white)(aaaaa)` `0` `color(white)(aaaaaaaa)` `2` `color(white)(aaaaaaaa)` `4` `color(white)(aaaaaa)` `+oo`

`color(white)(aaaa)` `t-2` `color(white)(aaaaaa)` `-` `color(white)(aa)` `0` `color(white)(aaa)` `+` `color(white)(aaaaaa)` `+`

`color(white)(aaaa)` `t-4` `color(white)(aaaaaa)` `-` `color(white)(aa)` ###color(white)(aaaa)#`-` `color(white)(aaa)` `0` `color(white)(aa)` `+`

`color(white)(aaaa)` `v(t)` `color(white)(aaaaaa)` `+` `color(white)(aaa)` `0` `color(white)(aaa)` `-` `color(white)(aaa)` `0` `color(white)(aa)` `+`

`color(white)(aaaa)` `s(t)` `color(white)(aaaaaa)` `↗` `color(white)(aa)` `20` `color(white)(aaa)` `↘` `color(white)(aa)` `16` `color(white)(aa)` `↗`

The sign chart fot the acceleration is as follows :

`color(white)(aaaa)` `"Interval"` `color(white)(aaaa)` `(0,3)` `color(white)(aaaa)` `(3,+oo)`

`color(white)(aaaa)` `a(t)` `color(white)(aaaaaaaaa)` `-` `color(white)(aaaaaaa)` `+`

`color(white)(aaaa)` `s(t)` `color(white)(aaaaaaaaa)` `nn` `color(white)(aaaaaaa)` `uu`

graph{(y-(x^3-9x^2+24x))(y-(3x^2-18x+24))(y-(6x-18))=0 [-23.9, 41.07, -6.7, 25.78]}