Kinematics-1D

When a particle is released from rest position under gravity, then $ v = 0 $, but $ a \neq 0 $.
Thus, Assertion is correct but Reason in incorrect.

To find value of time at which velocity is maximum, taking differentiation of $ v $ with respect to time
$\frac {d v}{d t} = 0$
Given, $v = 4t(1 - 2t)$
$v = 4 t - 8 t ^ {2} \Rightarrow \frac {d}{d t} (4 t - 8 t ^ {2}) = 0$
$\Rightarrow \quad 4 - 16t = 0 \Rightarrow t = \frac{1}{4} s = 0.25s$
Again, taking differentiation, we get
$\Rightarrow \quad \frac{d^2v}{dt^2} = -16 < 0$
So, at $ t = 0.25s $, velocity is maximum.

Time taken to reach the highest point from the height h is obtained from equation, $ v = u - gt $
$\therefore \quad 0 = u - g t \text { or } t = \frac {u}{g}$
(c: At highest point, final velocity of ball = 0)
Height attained above $ h $ is obtained from $ v^2 - u^2 = 2(-g)h $
or $0 - u^{2} = 2(-g)h_{1}$ or $h_1 = \frac{u^2}{2g}$
Total height, $ h_2 = \left[h_1 + h = \frac{u^2}{2g} + h\right] $
Time taken to hit the ground is obtained from
$h _ {2} = u t + \frac {1}{2} a t ^ {2} \text { or } \frac {u ^ {2}}{2 g} h + h = 0 + \frac {1}{2} g t ^ {2}$
∴ Total time taken, $ t = \sqrt{\frac{(u^{2} + 2gh)}{g}} $

Speed of walking, $ v_{1} = \frac{h}{h} $
Speed of escalator, $ v_{2} = \frac{h}{t_{2}} $
Time taken when she walks over moving escalator,
$\begin{array}{l} t = \frac {h}{v _ {1} + v _ {2}} \\ \Rightarrow \quad \frac {1}{t} = \frac {v _ {1}}{h} + \frac {v _ {2}}{h} = \frac {1}{t _ {1}} + \frac {1}{t _ {2}} \\ \Rightarrow \quad t = \frac {t t _ {2}}{t _ {1} + t _ {2}} \\ \end{array}$

For a car in motion, if we describe this event w.r.t. a frame of reference attached to the person sitting inside the car, the car will appear to be at rest as the person inside the car (i.e. observer) is also moving with same velocity and in the same direction as car.

$h$ 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, $t = \frac{h}{V_1 + V_2}$ $\frac{1}{t} = \frac{V_1 + V_2}{h} = \frac{1}{t_1} + \frac{1}{t_2} \Rightarrow t = \frac{t_1 t_2}{t_1 + t_2}$ .

Velocity of the particle is given as $ v = At + Bt^2 $
where, $A$ and $B$ are constants.
$\begin{array}{l} \Rightarrow \quad \frac {d x}{d t} = A t + B t ^ {2} \quad \left(\because v = \frac {d x}{d t}\right) \\ \Rightarrow \quad d x = (A t + B t ^ {2}) d t \\ \text { Integrating both sides, we get } \\ \int_ {x _ {1}} ^ {x _ {2}} d x = \int_ {1} ^ {2} A t + B t ^ {2} k t d t \\ \Rightarrow \quad \Delta x = x _ {2} - x _ {1} = A \int_ {1} ^ {2} t d t + B \int_ {1} ^ {2} t ^ {2} d t \\ = A \left[ t ^ {2} \right] _ {2} ^ {2} + B \left[ t ^ {3} \right] _ {3} ^ {2} \\ = \frac {A}{2} (2 t ^ {2} - 1 ^ {2}) + \frac {B}{3} (2 ^ {3} - 1 ^ {3}) \\ \end{array}$
∴ Distance travelled by the particle between 1s and 2s is
$\Delta x = \frac {A}{2} \times (3) + \frac {B}{3} (7) = \frac {3 A}{2} + \frac {7 B}{3}$

$V = At + Bt^2 \Rightarrow \frac{dx}{dt} = At + Bt^2$ $\Rightarrow \int_{0}^{x} dx = \int_{1}^{2} (At + Bt^2) dt$ $\Rightarrow x = \frac{A}{2}(2^2 - 1^2) + \frac{B}{3}(2^3 - 1^3) = \frac{3A}{2} + \frac{7B}{3}.$

Given, $ v = \beta x^{-2n} $
$a = \frac {d v}{d t} = \frac {d x}{d t} \cdot \frac {d v}{d x}$
$\Rightarrow \quad a = v \frac {d v}{d x} = \beta x ^ {- 2 n}) (- 2 \beta \beta x ^ {- 2 n - 1})$
$\Rightarrow \quad a = - 2 \beta^ {2} x ^ {2 - 4 n - 1}$

Let the ball hit water in $ t $ second.
For first ball, $ s = ut + \frac{1}{2} at^2 $

Given, upward velocity of ball = $ 25.2 \, ms^{-1} $
Height attained by the ball,
$H = \frac {u ^ {2}}{2 g} = \frac {2 5 . 2 \times 2 5 2}{2 \times 9 . 8} = 3 2. 4 \mathrm{m}$
Now, time taken by the ball to attain 32.4 m is
$t = \frac {u}{g} = \frac {2 5 2 . 3}{9 . 8} = 2 5 7 \mathrm{s}$

According to question, the given condition can be depicted as
$\begin{array}{c} \frac {u}{A} = \frac {\text {Mid - point}}{\text {C}} \\ \text {Time} t \text {——} \end{array}$
For motion from $A$ to $C$,
$\frac {s}{2} = 2 u t + \frac {1}{2} a \times (2 t) ^ {2} = 2 u t + 2 a t ^ {2} \quad \dots \tag {i}$
Also, for motion from C to B,
$\frac {s}{2} = v ^ {\prime} t + \frac {1}{2} a t ^ {2} = (u + 2 a t t + \frac {1}{2} a t ^ {2}$
(Let initial velocity be $ v' = u + 2at $)
$= u t + 2 a t ^ {2} + \frac {1}{2} a t ^ {2}$
Now, from Eq. (i), we get

At first, the slope is decreasing, therefore the motion is retarded. Finally, the displacement becomes constant, thus the particle stops.

Given, $ u = 0, v = 180 \mathrm{~kmh}^{-1} = 50 \mathrm{~ms}^{-1} $
Time taken, $ t = 10\mathrm{s} $, $ a = \frac{v - u}{t} = \frac{50}{10} = 5\mathrm{ms}^{-2} $
Distance covered by the car,
$\begin{array}{l} s = u t + \frac {1}{2} a t ^ {2} = 0 + \frac {1}{2} \times 5 \times (1 0) ^ {2} \\ = \frac {5 0 0}{2} = 2 5 0 \mathrm{m} \\ \end{array}$

The slope of velocity-time graph gives acceleration,
i.e. $\frac{\partial\underline{\lambda}}{\partial p} = \frac{\tan\theta_1}{\tan\theta_2}.$
Here, $\theta_{1}$ is the angle made by line $A$ with time axis and $\theta_{2}$ is the angle made by line $B$ with time axis
$\Rightarrow \quad \frac {\partial \underline {{\lambda}}}{\partial p} = \frac {\tan 2 5 ^ {\circ}}{\tan 5 0 ^ {\circ}}$

$v(x) = bx^{-2n}$ $a = v\frac{dv}{dx} = bx^{-2n}\{b(-2n)x^{-2n-1}\} = -2b^2 nx^{-4n-1}.$

A

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

$V_{A} = 10(-\hat{i})$
$V _ {B} = 1 0 (\hat {j})$
$V _ {B A} = 1 0 \hat {j} + 1 0 \hat {i}$
Time for shortest distance

$= \frac {1 0 0 / \sqrt {2}}{1 0 \sqrt {2}} = 5.$

$0 = - 5 + 1 0 t \Rightarrow \therefore t = 2. 5 \sec$
Hence, height $h = \frac{1}{2} \times 10 \times (2.5)^2 = 31.2m$ .

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

The position of an object is always expressed w.r.t. some reference point. If the initial position of an object w.r.t. a reference points is $\underline{s}$ and after sometime, it changes to $s_2$, then the magnitude of the displacement of the object is $s_2 - s_1$. Since, the given ball returns to its starting point in $10~s$, So, in $5s$, if it reaches the highest point which is at $s_2$.
Given, $ s_2 $ is at a height $ h $ w.r.t. $ s_2 $. Thus, the displacement after $ 5s $ will be $ s_2 - s_1 = h $.

Let $ v_{PG} = $ velocity of police w.r.t. ground,
$ v_{PG} = \text{velocity of thief w.r.t. ground,} $
$v_{IP} = \text{velocity of thief w.r.t. police} = v_{PG} - v_{PG}$
$= \left(\frac {1 5 3 - 4 5}{1 8}\right) \times 5 \approx 3 0 \mathrm{ms} ^ {- 1}$
$v_{PG}$ : velocity of bullet w.r.t. car
$= 1 8 0 - 3 0 = 1 5 0 \mathrm{ms} ^ {- 1}$

We know that, $ s_{\mathrm{ab}} = u + \frac{1}{2} a(2n - 1) $
$s _ {\mathrm{bd}} = 0 + \frac {1}{2} a (2 \times 3 - 1) = \frac {5}{2} a \quad (\text { For } n = 3 \mathrm{s})$
$s _ {\mathrm{fd} 1} = 0 + \frac {1}{2} a (2 \times 4 - 1) = \frac {7}{2} a \quad (\text { For } n = 4 \mathrm{s})$
So, the percentage increase $= \frac{s_{\mathrm{th}} - s_{\mathrm{bd}}}{s_{\mathrm{bd}}} \times 100$
$= \frac {\frac {7}{2} a - \frac {5}{2} a}{5} \times 100 = \frac {3}{2} a \times 100 = 2 \times 20 = 40 \% \tag{2}$

As from the question,

Average speed of the car, $ v_{av} $ ,
$= \frac {\text { Total distance travelled by the body } (d)}{\text { Total time taken } (t _ {\mathrm{f}} + t _ {2})} \quad \dots \tag {1}$
So, now time taken in 1st half of the distance,
$t _ {\mathrm{f}} = \text { Time } = \frac {\text { Distance }}{\text { Velocity }} = \frac {d}{2 \times 4 0}$
$\Rightarrow \quad t _ {\mathrm{f}} = \frac {d}{8 0} \quad (\because \text { Distance } = \frac {d}{2}, \text { velocity } = 4 0 \mathrm{km} ^ {- 1})$
Time taken for 2nd half, $ t_2 = \frac{d}{2 \times \text{Velocity}} $
So, $t_2 = \frac{d}{2\times 60} = \frac{d}{120}$
$(\because \text { distance } = \frac {d}{2}, \text { velocity } = 6 0 \mathrm{km} ^ {- 1})$
Now, average velocity, $ v_{\mathrm{av}} = \frac{d}{\frac{60}{80}} + \frac{d}{\frac{120}{200}} $ [From Eq. (i)]
$= \frac {d}{\frac {2 \times 8 0 \times 1 2 0}{2 0 0 d}} = 4 8 \mathrm{km} ^ {- 1}$