A particle leaves the origin with an lintial veloity ` vec u = (3 hat i)` and a constant acceleration ` vec a= (-1.0 hat i-0 5 hat j)ms^(-1)`. Its velocity ` vec v` and position vector ` vec r` when it reaches its maximum x-coordinate aer .

Options

A. `vec(v)=-2hat(j)`
B. `vec(v)=-1.5 hat(j) m//s`
C. `vec(r)=(4.5 hat(i)-2.25hat(j)) m`
D. `vec(r)=(3hat(i)-2hat(j)) m`

Answer

B, C

Explanation

`vec(v)(t)=(3-1xxt)hat(i)+(0-0.5t)hat(j)` …(i)
For maximum positive x coordinate when `v_(x)` becomes zero
`:. 3-t=0 rArrt=3` sec
then `vec(r)(3)=4.5hat(i)-2.25 hat(j)`