Kinematics-1D

Ans. (4)
v = u + a t
6 0 = 1 0 + 2 t
$t = 2 5 \mathrm{sec}.$
$\text { Correct option } (4)$

Ans. (2)
$R = \frac {u ^ {2}}{g} \sin 2 \theta$
R is maximum for $2\theta = 90^{\circ}$ .

Ans. (4)
$R _ {1} = \frac {u _ {1} ^ {2} \sin 2 \theta_ {1}}{g}; R _ {2} = \frac {u _ {2} ^ {2} \sin 2 \theta_ {2}}{g}$
$\frac {R _ {1}}{R _ {2}} = \frac {u _ {1} ^ {2}}{u _ {2} ^ {2}} \frac {\sin 2 \theta_ {1}}{\sin 2 \theta_ {2}} = \frac {4 0 ^ {2} \sin (2 \times 3 0 ^ {\circ})}{6 0 ^ {2} \sin (2 \times 6 0 ^ {\circ})} = \frac {4}{9}$

Ans. (A) $\vec{v} = \alpha(y\hat{x} + 2x\hat{y})$ $v_x = \alpha y \quad v_y = 2\alpha x$ $\frac{dv_x}{dt} = \alpha\frac{dy}{dt} = 2\alpha^2x \quad \frac{dv_y}{dt} = 2\alpha v_x = 2\alpha^2y$ ∴ $\vec{F} = m\vec{a} = 2m\alpha^2(x\hat{x} + y\hat{y})$

Ans. (4)
$\begin{array}{r l} & {R = \frac {V ^ {2} (2 \sin \theta \cos \theta)}{g}} \\ & {t = \frac {V \sin \theta}{g} \Rightarrow V = \frac {g t}{\sin \theta}} \\ & {\Rightarrow R = \frac {g ^ {2} t ^ {2}}{\sin^ {2} \theta}. \frac {2 \sin \theta \cos \theta}{g}} \\ & {\quad \tan \theta = \frac {2 g t ^ {2}}{R} = \frac {2 0 t ^ {2}}{R}} \\ & {\quad \cot \theta = \frac {R}{2 0 t ^ {2}}} \end{array}$

Ans. (1)
For $0 \leq x \leq 200$
v = m x + C
$v = \frac {1}{5} x + 1 0$
$a = \frac {v d v}{d x} = \left(\frac {x}{5} + 1 0\right) \left(\frac {1}{5}\right)$
$a = \frac {x}{2 5} + 2 \Rightarrow \text { Straight line till } x = 2 0 0$
For $x > 200$
$v = \mathrm{constant}$
$\Rightarrow a = 0$
So, final a-x graph will be as follows-


Hence most appropriate option will be (1).
Otherwise it would be BONUS.

Ans. (2)
Option (2) represent correct graph for particle moving with constant acceleration, as for constant acceleration velocity time graph is straight line with positive slope and x-t graph should be an opening upward parabola.

Ans. (3)
In 4 sec. 1 $^{st}$ drop will travel
⇒ $\frac{1}{2} \times (9.8) \times (4)^{2} = 78.4 \, m$ ∴ 2 $^{nd}$ drop would have travelled
⇒ 78.4 - 34.3 = 44.1 m
Time for 2 $^{nd}$ drop $\frac{1}{2}(9.8)t^{2} = 44.1$ t = 3 sec
∴ Each drop have time gap of 1 sec
∴ 1 drop per sec

Ans. (3)
$u = \sqrt {2 g h}$
Now, $S = \frac{h}{3}$ , $a = -g$
$S = u t + \frac {1}{2} a t ^ {2}$
${\frac {h}{3}} = \sqrt {2 g h} t + {\frac {1}{2}} (- g) t ^ {2}$
$t ^ {2} \left(\frac {g}{2}\right) - \sqrt {2 g h} t + \frac {h}{3} = 0$
From quadratic equation
$t _ {1}, t _ {2} = \frac {\sqrt {2 g h} \pm \sqrt {2 g h - \frac {4 g}{2} \frac {h}{3}}}{g}$
$\frac {t _ {1}}{t _ {2}} = \frac {\sqrt {2 g h} - \sqrt {\frac {4 g h}{3}}}{\sqrt {2 g h} + \sqrt {\frac {4 g h}{3}}} = \frac {\sqrt {3} - \sqrt {2}}{\sqrt {3} + \sqrt {2}}$

Ans. (4)
$\begin{array}{r} \vec {V} _ {B W} = 4 \sqrt {2} \cos 4 5 \hat {\imath} + 4 \sqrt {2} \sin 4 5 \hat {j} \\ = 4 \hat {\imath} + 4 \hat {j} \end{array}$
$\vec {V} _ {W} = - \hat {j}$
$\vec {V} _ {B} = \vec {V} _ {B W} + \vec {V} _ {W} = 4 \hat {\imath} + 3 \hat {j}$
$\vec {S} _ {B} = \vec {V} _ {B} \times t = (4 \hat {\imath} + 3 \hat {j}) \times 3 = 1 2 \hat {\imath} + 9 \hat {j}$
$\left| \vec {S} _ {B} \right| = \sqrt {(1 2) ^ {2} + (9) ^ {2}} = 1 5 m$

Ans. (4)
Rain is falling vertically downwards.
$\vec {v} _ {r / m} = \vec {v} _ {r} - \vec {v} _ {m}$
$\tan 6 0 ^ {\circ} = \frac {v _ {r}}{v _ {m}} = \sqrt {3}$
$v _ {r} = v _ {m} \sqrt {3} = v \sqrt {3}$
Now, $v_{m} = (1 + B)v$
and $\theta = 45^{\circ}$

$\tan 4 5 = \frac {v _ {r}}{v _ {m}} = 1$
$v _ {r} = v _ {m}$
$v \sqrt {3} = (1 + \beta) v$
$\sqrt {3} = 1 + \beta$
$\Rightarrow \beta = \sqrt {3} - 1 = 0. 7 3$

Ans. (3)
$x = u _ {x} t + \frac {1}{2} a _ {x} t ^ {2}$
$y = u _ {y} t + \frac {1}{2} a _ {y} t ^ {2}$
$3 2 = 0 \times t + \frac {1}{2} (4) (t) ^ {2}$
$t ^ {2} = 1 6$
$t = 4 \sec$
$\begin{array}{r l} & x = 3 \times 4 + \frac {1}{2} \times 6 \times 4 ^ {2} \\ & \quad = 1 2 + 4 8 = 6 0 m \\ & \therefore \quad \text {Correct answer (3).} \end{array}$

Ans. (3)
Given $\vec{u}=5\hat{j}m/s,\vec{a}=10\hat{i}+4\hat{j}$ , final coordinate $(20,y_{0})$ in time t
$S _ {x} = u _ {x} t + \frac {1}{2} a _ {x} t ^ {2}$
$2 0 - 0 = 0 + \frac {1}{2} \times 1 0 \times t ^ {2}$
$t = 2 \sec$
$S _ {y} = u _ {y} \times t + \frac {1}{2} a _ {y} t ^ {2}$
$y _ {0} = 5 \times 2 + \frac {1}{2} \times 4 \times 2 ^ {2} = 1 8 m$
2 sec and 18 m.

Ans. (ABCD)

$y = \frac{x^2}{2}$ At $t = 0, \left. \begin{array}{l} x = 0, y = 0 \\ u = 1 \end{array} \right\}$ given $y = \frac{x^2}{2}$ $\frac{dy}{dt} = \frac{1}{2}.2x\frac{dx}{dt} \Rightarrow v_y = xv_x$ Differentiate w.r.t. time $a_y = \frac{dx}{dt}.V_x + xa_x$ $a_y = v_x^2 + xa_x$ Option:
(A) If $a_x = 1$ and particle is at origin
(x = 0, y = 0) $a_y = v_x^2$ $a_y = 1^2 = 1$ At origin, at t = 0 sec
Speed = 1 (given)
(B) Option $a_y = v_x^2 + xa_x$ Given in option B, $a_x = 0$ $\Rightarrow a_y = v_x^2$ If $a_x = 0, v_x = \text{constant} = 1$ , (all the time) $\Rightarrow a_y = 1^2 = 1$ (all the time)
(C) At $t = 0, x = 0, v_y = xv_x$ Speed = 1 $v_y = 0$ $v_x = 1$ (D) $a_y = v_x^2 + xa_x$ $v_y = xv_x$ $a_x = 0$ (given in D option) $\Rightarrow a_y = v_x^2$ If $a_x = 0 \Rightarrow V_x = \text{constant initially } (v_x = 1)$ $\Rightarrow a_y = 1^2 = 1$ At $t = 1 \sec$ $v_y = 0 + a_y × t = 1 × 1 = 1$ $\tan \theta = \frac{v_y}{v_x} = x$ (θ → angle with x axis) $\tan \theta = \frac{v_y}{v_x} = \frac{1}{1} = 1$ θ = 45°

Ans. (2)
$v = b \sqrt {x}$
${\frac {d \nu}{d t}} = {\frac {b}{2 {\sqrt {x}}}} {\frac {d x}{d t}}$
$a = \frac {b \nu}{2 \sqrt {x}}$
$a = \frac {b (b \sqrt {x})}{2 \sqrt {x}}$
${\frac {d \nu}{d t}} = a = {\frac {b ^ {2}}{2}}$
$v = \frac {b ^ {2}}{2} \tau$

Ans. (3)
$v _ {z} = a \omega$
$v = \sqrt {2} a \omega$

Ans. (4)
$R = \frac {u ^ {2} \sin 2 \theta}{g}$
$A = \pi R ^ {2}$
$A \propto R ^ {2}$
$A \propto u ^ {4}$
${\frac {A _ {1}}{A _ {2}}} = {\frac {u _ {1} ^ {4}}{u _ {2} ^ {4}}} = \left[ {\frac {1}{2}} \right] ^ {4} = {\frac {1}{1 6}}$

Ans. (1)
$\begin{array}{r l} & {\vec {S} = (5 \hat {\imath} + 4 \hat {j}) 2 + \frac {1}{2} (4 \hat {\imath} + 4 \hat {j}) 4} \\ & {\quad = 1 0 \hat {\imath} + 8 \hat {j} + 8 \hat {\imath} + 8 \hat {j}} \end{array}$
$\vec {r} _ {f} - \vec {r} _ {i} = 1 8 \hat {\imath} + 1 6 \hat {j}$
$\vec {r} _ {f} = 2 0 \hat {\imath} + 2 0 \hat {j}$
$\left| \vec {r} _ {f} \right| = 2 0 \sqrt {2}$

Ans. (2)
$t = \frac {2 \times 2 \times \sin 1 5 ^ {\circ}}{g \cos 3 0 ^ {\circ}}$
$S = 2 \cos 1 5 ^ {\circ} \times t - \frac {1}{2} g \sin 3 0 ^ {\circ} t ^ {2}$
Put values and solve
$S \simeq 2 0 c m$

Ans. (1)
Equation of trajectory is given as
$y = 2 x - 9 x ^ {2}\tag{... (1}$
Comparing with equation :
$y = x \tan \theta - \frac {g}{2 u ^ {2} \cos^ {2} \theta}. x ^ {2}$
... (2)
We get;
$\tan \theta = 2$
$\therefore \qquad \boxed {\cos \theta = \frac {1}{\sqrt {5}}}$
$\begin{array}{r l} & {\mathrm{Also}, \frac {\mathsf {g}}{2 \mathsf {u} ^ {2} \cos^ {2} \theta} = 9} \\ & {\Rightarrow \quad \frac {1 0}{2 \times 9 \times \left(\frac {1}{\sqrt {5}}\right) ^ {2}} = u ^ {2}} \\ & {\Rightarrow \quad u ^ {2} = \frac {2 5}{9}} \\ & {\Rightarrow \quad \boxed {u = \frac {5}{3} m / s}} \end{array}$

Ans. (4)
$t _ {2} = \frac {1 2 0 + 6 0}{1 1 0 \times \frac {5}{1 8}}$
$t _ {1} = \frac {1 2 0 + 6 0}{5 0 \times \frac {5}{1 8}}$
$\frac {t _ {1}}{t _ {2}} = \frac {1 1 0}{5 0} = \frac {1 1}{5} \quad \text { Ans. }$

Ans. (4)
$\vec {r} = 1 5 t ^ {2} \hat {\imath} + (4 - 2 0 t ^ {2}) \hat {j}$
$\vec {v} = \frac {d \vec {r}}{d t} = 3 0 t \hat {\imath} + (- 4 0 t) \hat {j}$
$\vec {a} = \frac {d \vec {v}}{d t} = 3 0 \hat {\imath} - 4 0 \hat {j}$
$| \vec {a} | = 5 0 m / s ^ {2}$

The arrangement of the ships is shown in the Figure (not drawn to scale). Given that

$\vec {v} _ {A} = 3 0 \hat {i} + 5 0 \hat {j} \mathrm{kmhr} ^ {- 1}, \vec {r} _ {A} = 0 \hat {i} + 0 \hat {j}$
$\vec {v} _ {B} = \left(- 1 0 \hat {i}\right) \mathrm{kmhr} ^ {- 1}, \vec {r} _ {B} = \left(8 0 \hat {i} + 1 5 0 \hat {j}\right) \mathrm{km}$
$\Rightarrow \vec {r} _ {B A} = (8 0 \hat {i} + 1 5 0 \hat {j}) \mathrm{km}$
$\Rightarrow \vec {v} _ {B A} = \vec {v} _ {B} - \vec {v} _ {A} = - 1 0 \hat {i} - 3 0 \hat {i} - 5 0 \hat {i} = - 4 0 \hat {i} - 5 0 \hat {j}$
The time after which distance between the ships is minimum is given by
$t = \frac {\vec {r} _ {B A} \cdot \vec {v} _ {B A}}{\left| \vec {v} _ {B A} \right| ^ {2}}$
$\Rightarrow t = \frac {\left(8 0 \hat {i} + 1 5 0 \hat {j}\right) \cdot \left(- 4 0 \hat {i} - 5 0 \hat {j}\right)}{\left| - 4 0 \hat {i} - 5 0 \hat {j} \right| ^ {2}}$
$\Rightarrow t = \left(\frac {3 2 0 0 + 7 5 0 0}{4 1 0 0}\right) \mathrm{hr}$
$\Rightarrow t = \frac {1 0 7 0 0}{4 1 0 0} \mathrm{hr} = 2. 6 \mathrm{hr}$
Hence, the correct answer is (D).