Kinematics-1D

Here, $\vec{r}_{1}(t)=3t\hat{i}+4t^{2}\hat{j}$ and $\vec{r}_{2}(t)=4t^{2}\hat{i}+3t\hat{j}$
Velocity, $\vec{v}_1(t) = \frac{d\vec{r}_1}{dt} = \frac{d}{dt} (3t\hat{i} +4t^2\hat{j}) = 3\hat{i} +8t\hat{j}$
$\vec {v} _ {2} (t) = \frac {d \vec {r} _ {2}}{d t} = \frac {d}{d t} (4 t ^ {2} \hat {i} + 3 t \hat {j}) = 8 t \hat {i} + 3 j$
The relative speed of particle 1 with respect to particle 2 is
$\vec {v} _ {1 2} = \vec {v} _ {1} - \vec {v} _ {2}$
$= (3 \hat {i} + 8 t \hat {j}) - (8 t \hat {i} + 3 \hat {j}) = (3 - 8 t) \hat {i} + (8 t - 3) \hat {j}$
At $t = 1s$
$\vec {v} _ {1 2} = (3 - 8) \hat {i} + (8 - 3) \hat {j} = - 5 \hat {i} + 5 \hat {j}$
$| \bar {v} _ {1 2} | = \sqrt {(- 5) ^ {2} + (5) ^ {2}} = \sqrt {2 5 + 2 5} = 5 \sqrt {2} m / s.$

$h = ut - \frac{1}{2} gt^2 \Rightarrow 96 = 80t - \frac{32}{2} t^2$
$\Rightarrow t^{2} - 5t + 6 = 0 \Rightarrow t = 2$ sec or 3 sec.

$\because \frac{x_1}{x_2} = \frac{t_1}{t_2}$ and $\alpha = \frac{0 - v}{t_1} \Rightarrow t_1 = \frac{-v}{\alpha}$
$\beta = \frac {0 - v}{t _ {2}} \Rightarrow t _ {2} = \frac {- v}{\beta}$
$\therefore \frac {t _ {1}}{t _ {2}} = \frac {\beta}{\alpha}, \text { Hence } \frac {x _ {1}}{x _ {2}} = \frac {t _ {1}}{t _ {2}} = \frac {\beta}{\alpha}.$

C

B

$\sqrt{x} = t + 1$
Squaring both sides, we get, $x=(t+1)^{2}=t^{2}+2t+1$ Differentiating it w.r.t time t, we get, $\frac{dx}{dt}=2t+2$ Velocity, $v=\frac{dx}{dt}=2t+2$ .

Let the man will be able to catch the bus after t s, then

Here, $u = 10ms^{-1}$ , $t = 2s$ , $S = 20m$ Using $S = ut + \frac{1}{2} at^2$ $\therefore 20 = 10 \times 2 + \frac{1}{2} \times a \times 2^2$ $0 = 2a \Rightarrow a = 0$ .

$\int_{6.25}^{0}\frac{dv}{\sqrt{v}} = -2.5\int_{0}^{t}dt$ $\left|2\sqrt{v}\right|_{6.25}^{0} = -2.5t$ $2\sqrt{6.25} = 2.5t$ $t = 2\mathrm{sec}.$

Let the stone falls through a height $h$ in $t \, s$ .
Here, $u = 0$ , $a = g$
Using $D_{n} = u + \frac{a}{2} (2n - 1)$
Distance travelled by the stone in the last second is
$\frac {9 h}{2 5} = \frac {g}{2} (2 t - 1)$
$[ \because u = 0 ]\tag{... (i}$
Distance travelled by the stone in $t$ s is
$h = \frac {1}{2} g t ^ {2}$
$[ \because u = 0 ]\tag{... (ii}$
Divide (i) by (ii), we get
$\frac {9}{2 5} = \frac {(2 t - 1)}{t ^ {2}}$

$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 20} = 20m / s$ .

O is the observation point at the ground. A and B are the positions of aircraft for which $\angle AOB = 30^{\circ}$ . Time taken by aircraft from A to B is 10s
$\Delta AOB$
$\tan 3 0 ^ {\circ} = \frac {A B}{3 4 0 0}$
$AB = 3400\tan 30^{\circ} = \frac{3400}{\sqrt{3}} m$
∴ Speed of aircraft,
$v = \frac {A B}{1 0} = \frac {3 4 0 0}{1 0 \sqrt {3}} = 1 9 6. 3 m s ^ {- 1}.$

B

Average velocity = $\frac{Displacement}{Time interval}$ A particle moving in a given direction with non-zero velocity cannot have zero speed.
In general, average speed is not equal to magnitude of average velocity. However, it can be so if the motion is along a straight line without change in direction.

$\vec{r} = 3t\hat{i} - t^2\hat{j} + 4\hat{k}$ Velocity, $\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(3t\hat{i} - t^2\hat{j} + 4\hat{k}) = 3\hat{i} - 2t\hat{j}$ At $t = 5s \Rightarrow \vec{v} = 3\hat{i} - 10\hat{j}$ $|\vec{v}| = \sqrt{(3)^2 + (-10)^2} = \sqrt{9 + 100} = \sqrt{109} = 10.44ms^{-1}$ .

$x = \frac{1}{t + 5} \Rightarrow v = \frac{dx}{dt} = -\frac{1}{(t + 5)^2}$ Acceleration, $a = \frac{dv}{dt} = \frac{2}{(t + 5)^3} \Rightarrow a \propto (\text{velocity})^{3/2}$ .

D

Let $v_{w}$ be the velocity of water and $v_{b}$ be the velocity of motorboat in still water.
The distance x covered by the motorboat in moving downstream in 6 h is
$x = (\nu_ {b} + \nu_ {w}) \times 6\tag{... (i}$
Same distance x covered by the motorboat in moving upstream in 10 h is
$x = (\nu_ {b} - \nu_ {w}) \times 1 0\tag{... (ii}$
Equating (i) and (ii), we get
$(\nu_ {b} + \nu_ {w}) \times 6 = (\nu_ {b} - \nu_ {w}) \times 1 0$
$\upsilon_ {w} = \frac {\upsilon_ {b}}{4} \quad \therefore x = (\upsilon_ {b} + \upsilon_ {w}) \times 6 = 7. 5 \upsilon_ {b}$
Time taken by the motorboat to cover the same distance in still water is
$t = \frac {x}{v _ {b}} = \frac {7 . 5 v _ {b}}{v _ {b}} = 7. 5 \text { hours. }$

B

Let two balls meet at depth h from platform
$h = \frac {1}{2} g (1 8) ^ {2} = v (1 2) + \frac {1}{2} g (1 2) ^ {2} \Rightarrow v = 7 5 m s ^ {- 1}$

Let the two bodies $A$ and $B$ respectively meet at a time $t$ , at a height $\frac{H}{2}$ from the ground.

Using $S = ut + \frac{1}{2} at^2$
For a body $A$ , $u = V_{0}, a = -g$ , $S = \frac{H}{2}$
$\therefore \frac {H}{2} = V _ {0} t - \frac {1}{2} g t ^ {2}\tag{... (i}$
For body $B$ , $u = 0$ , $a = +g$ , $S = \frac{H}{2}$
$\therefore \frac {H}{2} = \frac {1}{2} g t ^ {2}\tag{... (ii}$
Equating equations (i) and (ii), we get
$V _ {0} t - \frac {1}{2} g t ^ {2} = \frac {1}{2} g t ^ {2} \Rightarrow V _ {0} t = g t ^ {2} \text { or } t = \frac {V _ {0}}{g}$
Substituting the value of t in equation (i), we get
$\frac {H}{2} = V _ {0} \times \left(\frac {V _ {0}}{g}\right) - \frac {1}{2} g \left(\frac {V _ {0}}{g}\right) ^ {2} = \frac {V _ {0} ^ {2}}{g} - \frac {1}{2} \frac {V _ {0} ^ {2}}{g}$
$\frac {H}{2} = \frac {1}{2} \frac {V _ {0} ^ {2}}{g} \text { or } V _ {0} ^ {2} = g H \Rightarrow V _ {0} = \sqrt {g H}.$

Free fall of an object (in vacuum) is a case of motion with uniform acceleration.

$\begin{array}{r l} & h = u t - \frac {1}{2} g t ^ {2} \\ & = 1 0 \times 1 - \frac {1}{2} \times 1 0 \times 1 \\ & = 1 0 - 5 = 5 m. \end{array}$

$h = \frac{1}{2} gt^2$
$h ^ {\prime} = \frac {1}{2} g (t - t _ {0}) ^ {2}$
$h - h ^ {\prime} = \frac {1}{2} g [ t ^ {2} - (t - t _ {0}) ^ {2} ]$
$= \frac {1}{2} g [ t ^ {2} - t ^ {2} - t _ {0} ^ {2} + 2 t t _ {0} ]$
$\Delta h = \frac {1}{2} g t _ {0} (2 t - t _ {0})$
$\Delta h$ is increasing with time.

B