A particle starts from the origin at $t=0$ $s$ with a velocity of $10.0 \hat{ j } \;m / s$ and moves in the $x-y$ plane with a constant acceleration of $(8.0 \hat{ i }+2.0 \hat{ j }) \;m \,s ^{-2} .$

$(a)$ At what time is the $x$ - coordinate of the particle $16\; m ?$ What is the $y$ -coordinate of the particle at that time?

$(b)$ What is the speed of the particle at the time?

$(a)$ Velocity of the particle $=10 \widehat{j} \,m / s$ Acceleration of the particle $=(8.0 \hat{\imath}+2.0 \hat{\jmath}) \,m\,s ^{-2}$

$ \vec{a}=\frac{d \vec{v}}{d t}=8.0 \hat{ i }+2.0 \hat{ j }$

$d \vec{v}=(8.0 \hat{ i }+2.0 \hat{ j }) d t$

Integrating both sides:

$\vec{v}(t)=8.0 t \hat{ i }+2.0 t \hat{ j }+\vec{u}$

Where,

$u=$ Velocity vector of the particle at $t=0$

$v=$ Velocity vector of the particle at time $t$ But, $\vec{v}=\frac{d \vec{r}}{d t}$

$d \vec{r}=\vec{v} d t=(8.0 t \hat{ i }+2.0 t \hat{ j }+\vec{u}) d t$

Integrating the equations with the conditions: at $t=0 ; r=0$ and at $t=t ; r=r$ 

$\vec r=\vec{u} t+\frac{1}{2} 8.0 t^{2} \hat{ i }+\frac{1}{2} \times 2.0 t^{2} \hat{ j }$

$=\overrightarrow{u t}+4.0 t^{2} \hat{ i }+t^{2} \hat{ j }$

$=(10.0 \hat{ j }) t+4.0 t^{2} \hat{ i }+t^{2} \hat{ j }$

$x \hat{ i }+y \hat{ j }=4.0 t^{2} \hat{ i }+\left(10 t+t^{2}\right) \hat{ j }$

since the motion of the particle is confined to the $x$ $-y$ plane, on equating the coefficients of $\vec{\imath}$ and $\vec{\jmath},$ we get:

$x=4 t^{2}$

$t=\left(\frac{x}{4}\right)^{\frac{1}{2}}$

And $y=10 t+t^{2}$

When $x=16 \,m$ $t=\left(\frac{16}{4}\right)^{\frac{1}{2}}=2 s$

$\therefore y=10 \times 2+(2)^{2}=24 \,m$

$(b)$ Velocity of the particle is given by:

$\vec{v}(t)=8.0 t \hat{ i }+2.0 t \hat{ j }+\vec{u}$

at $t=2 \,s$

$\vec{v}(t)=8.0 \times 2 \hat{ i }+2.0 \times 2 \hat{ j }+10 \hat{ j }$

$=16 \hat{ i }+14 \hat{ j }$

$\therefore$ Speed of the particle

$\vec{v}| =\sqrt{(16)^{2}+(14)^{2}}=\sqrt{256+196}=\sqrt{452}$

$=21.26\; m / s$