Answer 4.21:

a) Velocity of the particle = 10 𝑗̂ 𝑚/𝑠
Acceleration of the particle = (8.0𝑖̂+ 2.0𝑗̂) ${\mathrm{ms}}^{-2}$Also,

$\mathrm{But},<mspace\; width="1em"></mspace>\begin{array}{l}\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=8.0\hat{i}+2.0\hat{j}\\ d\overrightarrow{v}=(8.0\hat{i}+2.0\hat{j})dt\end{array}$
Integrating both sides:

$\overrightarrow{v}\left(t\right)=8.0t\hat{i}+2.0t\hat{j}+\overrightarrow{u}$
Where,
$\overrightarrow{\mathrm{u}}$ = Velocity vector of the particle at t = 0
$\overrightarrow{\mathrm{v}}$ = Velocity vector of the particle at time t

$\begin{array}{l}\text{\hspace{0.17em}But,\hspace{0.17em}}\overrightarrow{v}=\frac{dr}{dt}\\ d\overrightarrow{r}=\overrightarrow{v}dt=(8.0t\hat{i}+2.0t\hat{j}+\overrightarrow{u})dt\end{array}$

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

$\begin{array}{l}\overrightarrow{r}=\overrightarrow{u}t+\frac{1}{2}8.0t^2\hat{i}+\frac{1}{2}\times 2.0t^{2}j\\ =\overrightarrow{u}t+4.0t^2\hat{i}+t^2\hat{j}\\ =\left(10.0\hat{j}\right)t+4.0t^2\hat{i}+t^2\hat{j}\\ x\hat{i}+y\hat{j}=4.0t^2\hat{i}+(10t+t^2)\hat{j}\end{array}$

Since the motion of the particle is confined to the x-y plane, on equating the coefficients of 𝑖⃗ and 𝑗⃗, we get:

$\begin{array}{l}x=4t^2\\ t=(\frac{x}{4}{)}^{\frac{1}{2}}\\ \text{\hspace{0.17em}And\hspace{0.17em}}y=10t+t^2\end{array}$

When x = 16 m:

$t=(\frac{16}{4}{)}^{\frac{1}{2}}=2s$

$\therefore$ y = 10 × 2 + ${\left(2\right)}^2$ = 24 m

b) Velocity of the particle is given by:

$\begin{array}{l}\overrightarrow{v}\left(t\right)=8.0t\hat{i}+2.0t\hat{j}+\overrightarrow{u}\\ \text{\hspace{0.17em}at\hspace{0.17em}}t=2s\\ \overrightarrow{v}\left(t\right)=8.0\times 2\hat{i}+2.0\times 2\hat{j}+10\hat{j}\\ =16\hat{i}+14\hat{j}\\ \therefore \text{\hspace{0.17em}Speed of the particle:\hspace{0.17em}}\\ \begin{array}{c}\left|\overrightarrow{v}\right|=\sqrt{(16{)}^2+(14{)}^2}\\ =\sqrt{256+196}=\sqrt{452}\\ =21.26m/s\end{array}\end{array}$