A particle is moving along a straight line OX, At a time t (in seconds) the distance x (in metres) of particle from point O is given by `x=10+6t-3t^(2)`. How long would the particle travel before coming to rest?

Answer

`(A) rarr (Q, S); (B) rarr (T); (C) rarr (R); (D) rarr (R)`

Explanation

For (A) : `v_(av)=sqrt((v_(avx))^(2)+(v_(avy))^(2))=sqrt((10 cos 30^(@))^(2)+((10 sin 30^(@)+0)/2)^(2))=sqrt(75+25/4)=5/2sqrt(13) m//s`
Angle with horizontal `theta=tan^(-1) (v_(avy)/v_(avx))=tan^(-1)((5//2)/(5sqrt(3)))=tan^(-1)(1/(2sqrt(3)))`
For (B) : By using `vec(v)=vec(u)+vec(a)t` we have `u/(g t)=sin 30^(@)rArr t=10/((10)(1//2))=2`
For (C) : Horizonatl range (R)`=(u^(2) sin 2theta)/(g)=(100xxsqrt(3)//2)/(10)=5sqrt(3) m`
For (D) : Change in linear momentum `=m u_(y)=sqrt(3)xx10 sin 30^(@)=5sqrt(3) N-s`