rrect Option is (A)

Given that,

x = a cos $\omega$t

y = a sin $\omega$t

z = a $\omega$t

Velocity in x-direction,

V_x = $\frac{dx}{dt}=-a\omega \sin \omega t$

Velocity in y-direction,

V_y = $\frac{dy}{dt}$ = a $\omega$cos $\omega$t

Velocity in z-direction,

V_z = $\frac{dz}{dt}$ = a$\omega$

Net velocity,

$\overrightarrow{V}$ = V_x$\hat{i}$ + V_y$\hat{j}$ + V_z$\hat{k}$

Speed = $\left|\overrightarrow{V}\right|=\sqrt{V_x^2+V_y^2+V_z^2}$

$=\sqrt{a^2{\omega }^2{\sin }^2\omega t+a^2{\omega }^2{\cos }^2\omega t+a^2{\omega }^2}$

$=\sqrt{a^2{\omega }^2\left({\sin }^2\omega t+{\cos }^2\omega t\right)+a^2{\omega }^2}$

$=\sqrt{2a^2{\omega }^2}$

$=\sqrt{2}a\omega$

rrect Option is (C)

Given,

$\overrightarrow{v}=K\left(y\hat{i}+x\hat{j}\right)$

$\therefore$   Velocity in x direction,

$v_x=\frac{dx}{dt}=Ky$   . . . . . (1)

Velocity in y direction,

v_y = $\frac{dy}{dt}$ = Kx . . . . . . . (2)

$\therefore$   $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{Kx}{Ky}$

$\Rightarrow$   $\frac{dy}{dx}=\frac{x}{y}$

$\Rightarrow$   ydy $=$ xdx

Integrating both sides we get,

$\int ydx=\int xdx$

$\Rightarrow$   $\frac{y^2}{2}=\frac{x^2}{2}+c$

$\Rightarrow$   $y^2=x^2+2c$

$\therefore$   General equation,

y² = x² + constant.

rrect Option is (D)

$\overrightarrow{v}=k\left(y\hat{i}+x\hat{j}\right)$ ........(1)

Also $\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}$

$\overrightarrow{v}=\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}$ ........(2)

Equating (1) and (2), we get

$\frac{dx}{dt}=ky$ .......(3)

and $\frac{dy}{dt}=kx$ ......(4)

Dividing (3) and (4), we get

$\frac{dy}{dx}=\frac{x}{y}$

$\Rightarrow ydy=xdx$

Integrating both sides of above equation, we get

$\int ydy=\int xdx$

$\Rightarrow y^2=x^2+$ constant

rrect Option is (A)

Given $\overrightarrow{u}=3\hat{i}+4\hat{j},\overrightarrow{a}=0.4\hat{i}+0.3\hat{j},t=10s$

$\overrightarrow{v}=\overrightarrow{u}+\overrightarrow{a}t$

$=3\hat{i}+4\hat{j}+\left(0.4\hat{i}+0.3\hat{j}\right)\times 10$

$=7\hat{i}+7\hat{j}$

We know speed is equal to magnitude of velocity.

$\therefore$ $\left|\overrightarrow{v}\right|=\sqrt{7^2+7^2}=7\sqrt{2}$ units

rrect Option is (C)

Average acceleration

$=\frac{changeinvelocioty}{time\mathrm{int}erval}$

$=\frac{\mathrm{\Delta }\overrightarrow{v}}{t}$

${\overrightarrow{v}}_1=+5\hat{i}$, towards east direction.

$\overrightarrow{v_2}=+5\hat{j}$, towards north direction

$\mathrm{\Delta }\overrightarrow{v}=\overrightarrow{v_2}-\overrightarrow{v_1}$ = $5\hat{i}-5\hat{j}$

$\therefore$ $\overrightarrow{a}=\frac{5\hat{j}-5\hat{i}}{10}=\frac{\hat{j}-\hat{i}}{2}$

$\therefore$ $a=\frac{\sqrt{1^2+{\left(-1\right)}^2}}{2}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}m{s}^{-2}$

$\tan \theta =\frac{v_2}{v_1}=\frac{5}{5}=1$

$\therefore$ $\theta ={45}^{\circ }$

Therefore the direction is North-west.