Determine

(a) the velocity when x = 3m

(b) the position when the velocity is again zero

(c) the position when the velocity is maximum.

$$a= 30-4.5x^2$$

Substituting $a = v dv/dx \\ vdv/dx = 30-4.5x^2 \\ v dv = (30-4.5x^2)dx$

Integrating both the sides

$v^2/2=30x-4.5x^3/3 + c \\ v^2/2=30x-1.5x^3+ c $

Put v= 0 at x=0 (given) to get value of c.

c is found to be 0

so,

$v^2/2=30x-1.5x^3 \\ i) v\space at\space x=3 \\ V^2= 60x-3x^3 \\ = 60(3)-3(27) \\ =180-81=99 \\ v= 9.95 m/s \\ ii) x\space at\space v=0 \\ 0= 60x-3x^3 \\ 60=3x^2 \\ x=4.47m$

iii) x at v maximum

$vdv/dx = 30-4.5x^2 \space\space\space (given) $

When v is maximum, $dv/dx=0 \\ 0 = 30 - 4.5x^2 \\ x= 2.58m$