The velocity vector of a body is given by `vec(v)=2hati+(3-6t)hatj m//s`. At the initial moment the body is at origin. Find x-coordinate (in m) at the time when its y-coordinate is maximum.

Answer

1

Explanation

`vec(v) =2hati+(3-6t)hatj`
x=2t+C`" " `[t=0,x=0]
`rArr x=2t......(i)`
`y=3t-3t^(2)+C`
`y=3t-3t^(2)......(ii)`
Put `t=x/2` inequation (2)
`y=(3x)/2 3(x/2)^(2)`
`y=(3x)/2-(3x^(2))/4`
`y=3/2(x-(x^(2))/2)`
For `y_(max)`
`(dy)/(dx)=0`
`(dy)/(dx)=3/2 (1-x)=0 " " rArr x=1m`