Kinematics-1D

$\vec{r} = 3t\hat{i} - t^2\hat{j} + 4\hat{k}$ 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

$(v')^2 = u^2 + 2ad$
$v ^ {2} = (v ^ {\prime}) ^ {2} + 2 a d$
$\text { Solving, we get }$
$v ^ {\prime} = \sqrt {\frac {v ^ {2} + u ^ {2}}{2}}.$

... (i)

I is not possible because total distance covered by a particle increases with time. II is not possible because at a particular time, position cannot have two values. III is not possible because at a particular time, velocity cannot have two values. IV is not possible because speed can never be negative.
IV is not possible because speed can never be negative.

B

As displacement is either smaller or equal to distance but never be greater than distance.

The velocity of balloon at height $h$ , $v = \sqrt{2\left(\frac{g}{8}\right)h}$ When the stone released from this balloon, it will go upward with velocity $v = \frac{\sqrt{gh}}{2}$ (Same as that of balloon). In this condition time taken by stone to reach the ground
$t = \frac {v}{g} \left[ 1 + \sqrt {1 + \frac {2 g h}{v ^ {2}}} \right]$
$= \frac {\sqrt {g h} / 2}{g} \left[ 1 + \sqrt {1 + \frac {2 g h}{g h / 4}} \right]$
$= \frac {2 \sqrt {g h}}{g} = 2 \sqrt {\frac {h}{g}}.$

Let $t_{AB}, t_{BC}, t_{CD}$ and $t_{DA}$ be the time taken by the aeroplane to go from A to B, B to C, C to D and D to A respectively. $\therefore t_{AB} = \frac{1000 km}{250 km h^{-1}} = 4h$ $\therefore t_{BC} = \frac{1000 km}{500 km h^{-1}} = 2h$ $\therefore t_{CD} = \frac{1000 km}{200 km h^{-1}} = 5h$ $\therefore t_{DA} = \frac{1000 km}{100 km h^{-1}} = 10h$ Average speed $= \frac{\text{Total distance covered}}{\text{Total time taken}}$ $= \frac{1000 km + 1000 km + 1000 km + 1000 km}{t_{AB} + t_{BC} + t_{CD} + t_{DA}}$ $= 190.5 km h^{-1}$ .

i) The displacement of the man from $A$ to $E$ is $\Delta x = x_{2} - x_{1} = 7m - (-8m) = +15m$ directed in the positive $x$ -direction.
ii) The displacement of the man from E to C is $\Delta x = -3m - (7m) = -10m$ directed in the negative x-direction.
iii) The displacement of the man from B to D is $\Delta x = 3m - (-7m) = +10m$ directed in the positive x-axis.

$\vec{v} = \vec{u} +\vec{a} t\Rightarrow \vec{v} = 3\hat{i} +4\hat{j} +(0.4\hat{i} +0.3\hat{j})\times 10$ $v = 3\hat{i} +4\hat{j} +4\hat{i} +3\hat{j} = 7\hat{i} +7\hat{j}$ $\Rightarrow |\vec{v}| = 7\sqrt{2}$ units.

According to given relation acceleration $a = \alpha t + \beta$
As $a = \frac{dv}{dt}\Rightarrow \alpha t + \beta = \frac{dv}{dt}$
Since particle starts from rest, its initial velocity is zero i.e., At time t = 0, velocity = 0.
$\Rightarrow \int_ {0} ^ {v} d v = \int_ {0} ^ {t} (\alpha t + \beta) d t \Rightarrow v = \frac {\alpha t ^ {2}}{2} + \beta t.$

u = 0, v = 180 km $h^{-1} = 50 ms^{-1}$
time taken t = 10 s
$a = \frac {v - u}{t} = \frac {5 0}{1 0} = 5 m s ^ {- 2}$
∴ Distance covered $S = ut + \frac{1}{2} at^{2}$
$= 0 + \frac {1}{2} \times 5 \times (1 0) ^ {2} \approx \frac {5 0 0}{2} = 2 5 0 m.$

$\begin{array}{l} \text { Average speed } \bar {\nu} = \frac {\text { Total distance travelled }}{\text { Total time taken }} \\ = \frac {8 . 4 k m + 2 k m}{t _ {1} + t _ {2}} = \frac {1 0 . 4 k m}{\left(\frac {8 . 4 k m}{7 0 k m / h}\right) + \frac {1}{2} h} \\ = \frac {1 0 . 4 k m}{0 . 1 2 h + 0 . 5 h} = 1 6. 8 k m / h. \end{array}$

From figure, $V_{B}\sin \theta = V_{W}$
$\text { in } \theta = \frac {V _ {w}}{V _ {B}} = \frac {1}{2} \Rightarrow \theta = 3 0 ^ {\circ}$
$[ \because V _ {B} = 2 V _ {W} ]$
Time taken to cross the river,
$t = \frac {D}{V _ {B} \cos \theta} = \frac {D}{V _ {B} \cos 3 0 ^ {\circ}} = \frac {2 D}{V _ {B} \sqrt {3}}.$

Since $c >> v$ (negligible).

At time $t$
Velocity of $A, v_{A} = u - gt$ upward
Velocity of $B, v_{B} = gt$ downward
If we assume that height $h$ is smaller
$\begin{array}{c} B \bullet u _ {A} = 0 \\ \Bigg \downarrow \\ A \bullet u _ {A} = u \end{array} \Bigg \uparrow h$
than or equal to the maximum height reached by A, then at every instant $v_{A}$ and $v_{B}$ are in opposite directions.
$\therefore V _ {A B} = v _ {A} + v _ {B}$
= u - gt + gt [Speeds in opposite directions get added]