Kinematics-1D

Rain is falling vertically downwards
$\vec {v} _ {r / m} = \vec {v} _ {r} - \vec {v} _ {m}$
$\begin{array}{l l} \tan 6 0 ^ {\circ} = \frac {v _ {r}}{v _ {m}} = \sqrt {3} & \Bigg \downarrow \longrightarrow \vec {v} _ {m} \\ v _ {r} = v _ {m} \sqrt {3} = v \sqrt {3} & \Bigg \downarrow \vec {v} _ {r} \end{array}$


Now, $v_{m} = (1 + \beta)v$
and $\theta = 45^{\circ}$
$\tan 4 5 ^ {\circ} = \frac {v _ {r}}{v _ {m}} = 1$
$\tan \theta_{1} = a_{1} = \frac{V_{\max}}{t_{1}}$

$v^{2} = u^{2} + 2as$
$v ^ {2} = 0 + 2 \left(\frac {q E}{m}\right) x \Rightarrow v ^ {2} = \frac {2 q E}{m} x.$

S= area of graph=Total distance covered $=\frac{1}{2}\times4\left(1+\frac{13}{3}\right)+\frac{1}{2}\times2\times\frac{5}{3}=\frac{37}{3}m.$

For $A$ and $B$ let time taken by $A$ is $t_0$

Given,
$v _ {A} - v _ {B} = v = (a _ {1} - a _ {2}) t _ {0} - a _ {2} t\tag{... (i}$
$x _ {B} = x _ {A} = \frac {1}{2} a _ {1} t _ {0} ^ {2} = \frac {1}{2} a _ {2} (t _ {0} + t) ^ {2}$
$\Rightarrow \sqrt {a _ {1}}. t _ {0} = \sqrt {a _ {2}} (t _ {0} + t)$
$\Rightarrow (\sqrt {a _ {1}} - \sqrt {a _ {2}}) t _ {0} = \sqrt {a _ {2}}. t\tag{... (ii}$
putting $t_0$ in equation (i)
$\begin{array}{l} v = (a _ {1} - a _ {2}) \frac {\sqrt {a _ {2}} t}{\sqrt {a _ {1}} - \sqrt {a _ {2}}} - a _ {2} t \\ = (\sqrt {a _ {1}} + \sqrt {a _ {2}}) \sqrt {a _ {2}} t - a _ {2} t = \sqrt {a _ {1} a _ {2}} t + a _ {2} t - a _ {2} t \\ \Rightarrow v = \sqrt {a _ {1} a _ {2}} t. \end{array}$

$\theta = \omega t = \omega \frac{\pi}{2\omega} = \frac{\pi}{2}$
$\begin{array}{r} \vec {V} _ {A} - \vec {V} _ {B} = \omega R _ {1} (- \hat {i}) - \omega R _ {2} (- \hat {i}) \\ = \omega (R _ {2} - R _ {1}) \hat {i}. \end{array}$

For swimmer to cross the river straight
$\Rightarrow 4\sin \theta = 2$ or $\sin \theta = \frac{1}{2}\Rightarrow \theta = 30^{0}$
So, angle with direction of river flow = $90^{0} + \theta = 120^{0}$

$x = at + bt^2 - ct^3$
$v = \frac {d x}{d t} = a + 2 b t - 3 c t ^ {2}$
$a = \frac {d v}{d t} = 2 b - 6 c t = 0 \Rightarrow t = \frac {b}{3 c}$
$V _ {\left(a t t = \frac {b}{3 c}\right)} = a + 2 b \left(\frac {b}{3 c}\right) - 3 c \left(\frac {b}{3 c}\right) ^ {2} = a + \frac {b ^ {2}}{3 c}.$

$\vec{r}(t)=15t^{2}\hat{i}+(4-20t^{2})\hat{j}$
$\vec {v} = \frac {d \vec {r}}{d t} = 3 0 \hat {t i} + (- 4 0 t) \hat {j}$
$\vec {a} = \frac {d \vec {v}}{d t} = 3 0 \hat {i} - 4 0 \hat {j} \Rightarrow | \vec {a} | = 5 0 m / s ^ {2}.$

$-(g + \gamma v^2) = \frac{dv}{dt}$
$- d t = \frac {1}{\gamma} \left(\frac {d v}{\frac {g}{\gamma} + v ^ {2}}\right) \Rightarrow - \int_ {0} ^ {t} d t = \frac {1}{\gamma} \int_ {v _ {0}} ^ {0} \frac {d v}{\left(\sqrt {\frac {g}{\gamma}}\right) ^ {2} + v ^ {2}}$
$- t = - \frac {1}{\sqrt {\gamma g}} \tan^ {- 1} \left(\frac {v _ {0}}{\sqrt {\frac {g}{\gamma}}}\right) \Rightarrow t = \frac {1}{\sqrt {\gamma g}} \tan^ {- 1} \left(\sqrt {\frac {\gamma}{g}} v _ {0}\right).$

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

If we take the position of ship ‘A’ as origin then positions and velocities of both ships can be given as
$\begin{array}{r l} & {\vec {v} _ {A} = (3 0 \hat {i} + 5 0 \hat {j}) k m / h r} \\ & {\vec {v} _ {B} = - 1 0 \hat {i} k m / h r} \\ & {\vec {r} _ {A} = 0 \hat {i} + 0 \hat {j}} \\ & {\vec {r} _ {B} = (8 0 \hat {i} + 1 5 0 \hat {j}) k m} \end{array}$
Time after which
distance between them
will be minimum

$t = \frac {\vec {r} _ {B A} . \vec {v} _ {B A}}{| \vec {v} _ {B A} | ^ {2}}$
where $\vec{r}_{BA} = (80\hat{i} + 150\hat{j})km$
$\hat {v} _ {B A} = - 1 0 \hat {i} - (3 0 \hat {i} + 5 0 \hat {j})$
$\begin{array}{l} v _ {B A} = - 1 0 i - (3 0 i + 5 0 j) \\ (- 4 0 \hat {i} - 5 0 \hat {j}) k m / h r \\ \therefore t = - \frac {(8 0 \hat {i} + 1 5 0 \hat {j}) . (- 4 0 \hat {i} - 5 0 \hat {j})}{| - 4 0 \hat {i} - 5 0 \hat {j} | ^ {2}} \\ = \frac {3 2 0 0 + 7 5 0 0}{4 1 0 0} h r = \frac {1 0 7 0 0}{4 1 0 0} h r = 2. 6 h r s. \end{array}$

$S = \text{Area under graph} = \left(\frac{1}{2} \times 2 \times 2\right) + (2 \times 2) + (3 \times 1) = 9m.$

Given initial velocity $u = 0$ and acceleration is constant
$v = 0 + a t \Rightarrow v = a t$
also $x = 0(t) + \frac{1}{2} at^2\Rightarrow x = \frac{1}{2} at^2$



Graph (A); (B) and (D) are correct.

As in distance vs time graph slope is equal to speed in the given graph slope increase initially which is incorrect.

$h$ is height to reach from one floor to other floor speed of walking $= \frac{h}{t_1} = V_1$ speed of escalator $= \frac{h}{t_2} = V_2$ so time taken when escalator is walking and she is also walking on it, $t = \frac{h}{V_1 + V_2}$ $\frac{1}{t} = \frac{V_1 + V_2}{h} = \frac{1}{t_1} + \frac{1}{t_2} \Rightarrow t = \frac{t_1 t_2}{t_1 + t_2}$ .

$\overrightarrow{V_{BA}} = \overrightarrow{V_{B}} - \overrightarrow{V_{A}}$
$= (2 0 \times \cos 6 0 ^ {\circ} \hat {i} + 2 0 \sin 6 0 ^ {\circ} \hat {j}) - 1 0 \hat {i} = 1 0 \sqrt {3} \hat {j}$

$V_{A}^{2}=u^{2}-2g(-h)=u^{2}+2gh$ $V_{B}^{2}=u^{2}+2gh=u^{2}+2gh$ $V_{A}=V_{B}$ (Attain the same final velocity).

Time to reach the maximum height $t_{1}=u/g$ If $t_{2}$ be the time taken to hit the ground $-H=ut_{2}-\frac{1}{2}gt_{2}^{2}$ But $t_{2}=nt_{1}$ (given) $\Rightarrow-H=u\frac{nu}{g}-\frac{1}{2}g\frac{n^{2}u^{2}}{g^{2}}$ $\Rightarrow 2g H = nu^{2}(n - 2).$

$h_1 = \frac{1}{2} g(5)^2$ , $h_2 = \frac{1}{2} g(10)^2$ and $h_3 = \frac{1}{2} g(15)^2$ $\Rightarrow h_1 = \frac{h_2}{3} = \frac{h_3}{5}$ .

For train $A$ ,
$u_{A} = 72km / h = 72\times \frac{5}{18} m / s = 20m / s$ For train $B$ $u_{B} = 72km / h = 72\times \frac{5}{18} m / s = 20m / s$ $\therefore u_{BA} = u_B - u_A = 20m / s - 20m / s = 0m / s$ $a_{BA} = a_B - a_A = 1m / s^2 -0m / s^2 = 1m / s^2$ $t = 50s$ $S_{BA} = u_{BA}t + \frac{1}{2} a_{BA}t^{2} = 0\times 60 + \frac{1}{2}\times 1\times (50)^{2} = 1250m.$