A particle moves in the `x-y` plane under acceleration `vec(a) = 3hat(i) + 4hat(j) m//s^(2)`. If the initial velocities is `vec(u) = 6hat(i) + 8hat(j) m//s`. Find the velocity and the position vector of a particle after `2 s`.

Explanation

`vec(u) = 6hat(i) + 8hat(j) = u_(x)hat(i) + u_(y)hat(j)`
`u_(x) = 6 m//s, u_(y) = 8 m//s`
`vec(a) = 3hat(i) + 4hat(j) = a_(x)hat(i) + a_(y)hat(j)`
`a_(x) = 3 m//s^(2), a_(y) = 4 m//s^(2)`
`x` : `v_(x) = u_(x) + a_(x) t = 6 + 3 xx 2 = 12 m//s`
`x = u_(x) t + (1)/(2) a_(x) t^(2) = 6 xx 2 + (1)/(2) xx 3 xx (2)^(2) = 18 m`
`y` : `v_(y) = u_(y) + a_(y) t = 8 + 4 xx 2 = 16 m//s`
`y = u_(y) t + (1)/(2) a_(y) t^(2) = 8 xx 2 + (1)/(2) xx 4 xx (2)^(2) = 24 m`
After `2 s`,
Velocity `vec(v) = v_(x) hat(i) + v_(y)hat(j) = 12 hat(i) + 16 hat(j) m//s`
Position vector `vec( r) = x hat(i) + y hat(j) = 18 hat(i) + 24 hat(j) m`