Answer From Eq. $(4.34a)$ for $r _{0}=0$, the position of the particle is given by

$r (t)= v _{ o } t+\frac{1}{2} a t^{2}$

$=5.0 \hat{ i } t+(1 / 2)(3.0 \hat{ i }+2.0 \hat{ j }) t^{2}$

$=\left(5.0 t+1.5 t^{2}\right) \hat{ i }+1.0 t^{2} \hat{ j }$

$\text {Therefore, } \quad x(t)=5.0 t+1.5 t^{2}$

$y(t)=+1.0 t^{2}$

Given $x(t)=84 m , t=?$

$5.0 t+1.5 t^{2}=84 \Rightarrow t=6 s$

At $t=6 s , y=1.0(6)^{2}=36.0 m$

Now, the velocity $v =\frac{ d r }{ d t}=(5.0+3.0 t) \hat{ i }+2.0 t \hat{ j }$

At $t=6 s , \quad v =23.0 \hat{ i }+12.0 \hat{ j }$

speed $=| v |=\sqrt{23^{2}+12^{2}} \cong 26 m s ^{-1}$