A particle starts from the origin at `t= 0` with a velocity of `8.0 hat j m//s` and moves in the x-y plane with a constant acceleration of `(4.0 hat i + 2.0 hat j) m//s^2.` At the instant the particle's x-coordinate is 29 m, what are
(a) its y-coordinate and
(b) its speed ?

Answer

A, B, D

Explanation

`x=u_xt+1/2a_xt^2`
`:. 29=(0)(t)+1/2xx(4.0)t^2`
`:. t^2=sqrt14.5s^2 or t=3.8s`
(a) `y=u_yt+1/2a_yt^2`
`=(8)(3.8)+1/2xx2xx14.5`
`=44.9m~~45m`
(b) `v=u+at`
`=(8 hatj)+(4.0 hati+2.0 hatj)(3.8)`
`=(15.2 hati+15.6 hatj)`
`:. Speed =|v|=sqrt((15.2)^2+(15.6)^2)`
`=22m//s`