Motion In A Plane – JEE Main PhysicsPractice Questions & PYQs

The situation is depicted in the following figure.where, $V_{M R}=$ velocity of $man=12 \frac{km}{h}$ and $v_{R}=$ velocity of water flow in river $=6 \frac{km}{h}$ As, $v_{M R}$ should be along $C D$. $\Rightarrow \;\; v_{R}-v_{M R} \sin \theta =0$ $\Rightarrow \;\; 6-12 \sin \theta =0 \Rightarrow \sin \theta =\;\frac{6}{12}$ $\Rightarrow \;\; \sin \theta =\;\frac{1}{2}$ $\Rightarrow \;\; \theta =\sin ^{-1}(\;\frac{1}{2})=\sin ^{-1}(\sin 30^{\circ}) \;\; [ \because \sin 30^{\circ}=\;\frac{1}{2}]$ $\Rightarrow \;\; \theta =30^{\circ}$ $\therefore \;\; \alpha =90^{\circ}+\theta =90^{\circ}+30^{\circ}=120^{\circ}$ $\Rightarrow \;\; \alpha =120^{\circ}$

Given, horizontal velocity of helicopter, $u_{x}=v$ Height of the helicopter from ground = h The given condition can be drawn in the figure given below Here, M is the position of man. Let, time taken by food packet to reach ground from helicopter = t Acceleration due to gravity, $g=10 {\text{ms}}^{-2}$ Initial velocity of food packet along Y-axis, $u_{y}=0$ Since, $h=u_{y} t+\frac{1}{2} g t^{2}$ $\therefore t=\\sqrt{\frac{2 h}{g}}$ and distance travelled by helicopter in time t along horizontal direction $\Rightarrow x=u_{x} t$ $=v \\sqrt{\frac{2 h}{g}}$ Hence, distance of helicopter from man, $\text{H}=\\sqrt{h^{2}+x^{2}}=\\sqrt{h^{2}+(v \\sqrt{\frac{2 h}{g}})^{2}}$ $=\\sqrt{h^{2}+\frac{v^{2} 2 h}{g}}$

Rain is falling vertically downwards. ${{\vec{v}}}_{{r} / {m}}={{\vec{v}}}_{{r}}-{{\vec{v}}}_{{m}}$ $\tan 60°={{v}_{{r}}}/{{v}_{{m}}}=\sqrt{3}$ ${v}_{{r}}={v}_{{m}} \sqrt{3}={v} \sqrt{3}$ Now, ${v}_{{m}}=(1+{\text{B}}) {v}$ and $\theta =45°$ $\tan 45°={{v}_{{r}}}/{{v}_{{m}}}=1$ ${v}_{{r}}={v}_{{m}}$ ${v} \sqrt{3}=(1+\beta ) {v}$ $\sqrt{3}=1+\beta$ $\Rightarrow \beta =\sqrt{3}-1=0.73$

$\frac{{h}_{1}}{ {d}}=\tan 45°\Rightarrow {h}_{1}= {d}$ ......(1) $\frac{{h}_{1}+ {h}_{2}}{ {d}+2.464 {d}}=\tan 30°$ $\Rightarrow ({h}_{1}+ {h}_{2} ) \times \sqrt{3}=3.46 \ {d}$ $({h}_{1}+ {h}_{2} )= \frac{3.46 {d}}{\\sqrt{3}}$ $\Rightarrow {d}+ {h}_{2}= \frac{3.46 {d}}{\\sqrt{3}}$ ${h}_{2}= {d}$

$x = 10 + 8t - 3t^2$ $v_x = 8 - 6t$ $(v_x)_{t=i }= 2\hat{i}$ $y = 5 - 8t^3$ $v_y = - 24t^2$ $(v_y)_{t=1}=-24\hat{j}$Now $\text{sqrtv}=\sqrt{24^{2+}2^2}=\sqrt580$ $\therefore v=580 m^2/s^2$

Given ${\vec{u}}=5\hat{j} {m} / {s}, {\vec{a}}=10 { \hat{i}}+4\hat{j}$final coordinate $(20, \ {y}_{0} )$ in time $\ {t}$ ${S}_{ {x}}=4_{ {x}} {t}+ \frac{1}{2} {a}_{ {x}} {t}^{2}$ $20-0=0+ \frac{1}{2} \\times 10 \times {t}^{2}$ $\ {t}=2 \ {sec}$ $\ {S}_{ {y}}= {u}_{ {y}} \times {t}+ \frac{1}{2} {a}_{ {y}} {t}^{2}$ ${y}_{0}=5 \\times 2+ \frac{1}{2} 4 \\times 2^{2}=18 \ {m}$ 2 sec and $18 \ {m}$

$\vec{r}(t) = \cos\omega t\hat{i} + \sin \omega t\hat{j}$On diff. we get$\vec{v}=-\omega \sin\omega t\hat{i}+\omega \cos\omega t\hat{j}$ $\vec{a}=-\omega ^2\vec{r}$ $\vec{v}.\vec{r}=0$

$x=u_xt+1/2a_xt^2$ $y=u_yt+1/2a_yt^2$ $32=0\times t+1/2\times 4\times t^2$ $t=4\sec$ $x=3\times 4+1/2\times 6\times 4^2=60m$

to cross the river straight, $V_{\text{along} \; \text{the}\; \text{river}}$ = 0 $\ {V}_{ { \text{steam} }}= {V}_{ { \text{swimmer} }} \cos \theta$ $2=4 \cos \theta \Rightarrow\cos \theta =\ \frac{1}{2}$ $\Rightarrow \theta =60^{°}$ Direction w.r.t flow of river = 180°- 60° = 120°

Velocity of A w.r.t B is ${\vec{V}_{A, B}}=40 \vec{i}+50 \vec{j} {km} / {hr}$ ${\vec{r}_{B, A}}=80 \vec{i}+150 \vec{j}$ $t=\ \frac{|r| \cos \alpha }{|V|}= \frac{| \vec{V}.\vec{r}| }{| \vec{V}|^{2}}=2.6 h r$

$R= \frac{u^{2} \sin 2 \theta }{g}$ $h_{1}= \frac{u^{2} \sin ^{2} \theta }{2 g}, h_{2}=\ \frac{u^{2} \cos ^{2} \theta }{2 g}$ $h_{1} h_{2}=\ \frac{u^{4} \sin ^{2} \theta \cos ^{2} \theta }{4 g} \times \frac{4}{4}$ $h_{1} h_{2}$ $=\ \frac{R^{2}}{16}$

AB = $V_p$ × t BC = Vt cos 60° = $\frac{AB}{BC}$ $1/2$ = $\frac{V_p\times t}{Vt}$ $V_P$ = $V/2$

$R= \frac{2 u^{2} \sin \alpha \cos (\alpha +\beta )}{g \cos ^{2} \beta }$ Here $\;\;\;\;\=\theta =15 {\°}$ $\beta =\alpha =30 {\°}$ $R=\ \frac{2 \\times 4 \times \sin 15° \cos \ (45°)}{g \cos ^{2} 30°}$ $=20 \ {cm}$

$\tan \theta =2$ $\cos \theta _{0}=\ \frac{1}{\\sqrt{5}}$ $\theta _{0}=\cos ^{-1} (\ \frac{1}{\\sqrt{5}}\ )$ $y=2 x-9 x^{2}$ $\ \frac{g}{2 V_{0}^{2} \cos ^{2} \theta }=9$ ${V_{0}^{2} \cos ^{2} \theta =\ \frac{10}{18}} \;\;\;\; {=\ \frac{5}{9}}$ ${V_{0} \cos \theta =\ \frac{\\sqrt{5}}{3}}$ {V_{0}= {5}{3} $m$ / $s$

R = ${u^2sin2\theta }/g$ A = π$R^2$ A α $R^2$ A α $u^4$ $A_1/A_2$ = $u_1^4/u_2^4$ = $[1/2]^4$ = $1/{16}$

$\frac{dx}{dt}$ = ky , $\frac{dy}{dt}$ = kx Now , $\frac{dy}{dx}$ = $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ = $x/y$ \Rightarrow ydy = xdx Integrating both side $y^2$ = $x^2$ + c

$v_x$ = - aωsinωt \Rightarrow $v_y$ = aωcosωt $v_z$ = aω \Rightarrow v = $\sqrt{v_x^{2+}v_y^{2+}v_z^2}$ v = $\sqrt2$ aω

$\ \frac{2 u \sin \theta }{g}=t_{1}$ $\ \frac{2 u \cos \theta }{g}=t_{2}$ $t_{1} t_{2}= \frac{2 R}{g}$

Let $RM = x$. The distance $QR = QM - x$, so the time taken is:
$T = \frac{QM - x}{v} + \frac{\sqrt{x^2 + d^2}}{v/2} = \frac{QM - x}{v} + \frac{2\sqrt{x^2 + d^2}}{v}$
To minimize time, we differentiate with respect to $x$ and set to zero:
$\frac{dT}{dx} = -\frac{1}{v} + \frac{2}{v} \cdot \frac{2x}{2\sqrt{x^2 + d^2}} = 0$
$\frac{2x}{\sqrt{x^2 + d^2}} = 1 \implies 4x^2 = x^2 + d^2$
$3x^2 = d^2 \implies x = \frac{d}{\sqrt{3}}$
Thus, the distance $RM = \frac{d}{\sqrt{3}}$.

$x=t$ $y=2t-5t^2$Equation of trajectory is $y=2x-5x^2$