if position vector is `vecr =t^3−6t^2+3t+4m ` find the value of velocity and acceleration.

Explanation

`tan theta=10/(10sqrt(3))=1/sqrt(3)`
`theta=30^(@)`
`theta=tan^(-1)=(u_(AB_(y))/u_(AB_(x)))=tan^(-1)((5sqrt(3))/15)=30^(@)`
Here `u_(AB_(y))=u_(A_(y))-u_(B_(y))=0-(-10 sin 60^(@))=5sqrt(3) m//s`
`u_(AB_(x))=u_(A_(x))-u_(B_(x))=10-(-10 cos 60^(@))=15 m//s`
So `u_(AB)` is along AB particle will collide.
`u_(AB)=sqrt(15^(2)+(5sqrt(3))^(2))=10sqrt(3) m//s`
`t=s_(AB)/u_(AB)=sqrt(10^(2)+(10sqrt(3))^(2))/(10sqrt(3))=20/(10sqrt(3))`
`t=2/sqrt(3)` sec.