A paarticle starts from origin at ` t=0 ` with a vction `5.0` hat I m//s` and moves in x-y` plane under action of a force which produces a constant acceleration of `( 3.0 hat I + 2.0 ht j) m//s^(2)`.
(a) What is the y-cordinate of the particle at the instant its x-coordinate id `84 m ? (b) What is the speed of the particle at this time?

Explanation

Taking motion along x-direction,
We have `u_(x) 5.0 m//s , a_(x) =3.0 m//s^(2)`,
`x= 84 m, t=? , v_(x) =?`
As, `x_(x) t+ 1/2 a_(x) t^(2)`
:. 84 =5.0 t + 1/2 xx 3 xx t^(2)`
or `168 =10 t + 3 t^(2) or `13 t^(2) + 10 t-168 =0`
On solving, `t=6 s`
`v_(x) =u_(x) a_(x) t= 5.0 + 3 xx 6 =23 m//s`
Taking motion along y-direction, we have
` u_(y) =0, a_(y) =2.0 m//s^(2), y=?, t= s, u_(y) =?`
As, y=1/2 a_(y) t^(2) =1/2 xx 2 xx 6 ^(2) =36 =36 m`
`u_(y) =u_(y) +a_(y) t=o +2 xx 6 =12 ms^(-1)`
:. `v= sqrt v_(x)^(2) + v_(y)^(2) =sqrt (23^(2) +12^(2))`
`=25 .9 ms^(-1)`.