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.?

Question

$t>=0$

10814 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-21T10:33:28

See the explanation below

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]}