Kinematics-1D

The distance covered in 2 s, $ x = \frac{1}{2} a(2)^2 $ ($ \because u = 0 $)
= 2 a
The distance covered in next 2 s,
$y = \frac {1}{2} a (4) ^ {2} - \frac {1}{2} a (2) ^ {2} = 6 a$
Now, $x / y = \frac{2a}{6a} = \frac{1}{3}\Rightarrow y = 3x$

$A \to v_{A}$ (constant velocity)
$B\to a$ (constant acceleration)
$C$ be the fixed point at which both $A$ and $B$ meet in time $t$.
Now, $ s = v_{A}t $ and $ v_{B} = at $
$A \rightarrow v _ {A} = v$

$v_{AB} =$ Speed of car from $AtoB$
$A \xlongequal {\quad} D \xlongequal {\quad} B$
$v _ {A B} = 3 0 \mathrm{km} \mathrm{h} ^ {- 1}$
$v _ {B A} = \text { Speed of car from } B \text { to } A$
$v _ {B A} = 2 0 \mathrm{km} ^ {- 1}$
$v _ {A} = \frac {\Delta_ {A B} v _ {B A}}{v _ {A B} + v _ {B A}} = 2 4 \mathrm{kmh} ^ {- 1}$
$A \xlongequal {\quad} D \xlongequal {\quad} B$
$v _ {A B} = 3 0 \mathrm{kmh} ^ {- 1}$
$v _ {B A} = \text { Speed of car from } B \text { to } A$
$v _ {B A} = 2 0 \mathrm{km} ^ {- 1}$
$v _ {A B} = \frac {\Delta_ {A B} v _ {B A}}{v _ {A B} + v _ {B A}} = 2 2 4 \mathrm{kmh} ^ {- 1}$
$v _ {A B} = \frac {v _ {A B} v _ {B A}}{v _ {A B} + v _ {B A}} = 2 4 \mathrm{kmh} ^{- 1}$

$(0,0)\xrightarrow{A}\frac{x_1}{A}\xrightarrow{C}\frac{x_2}{A}\xrightarrow{B}$ $u = 0\longleftarrow \longrightarrow r2\longrightarrow \longrightarrow r2\longrightarrow$
$a = \text{constant}$
Let $ \Lambda $ be the original point.
From equation of motion,
$x = u t + \frac {1}{2} a t ^ {2} = \frac {1}{2} a t ^ {2} \quad (\because u = 0)$
For point $C$, $x_{1} = \frac{1}{2} at / 2t^{2}\Rightarrow x_{1} = \frac{1}{8} at^{2}$
Distance travelled by the body in t second is
$x _ {1} + x _ {2} = \frac {1}{2} a t t ^ {2} = 4 x _ {1}$
$x _ {2} = 4 x _ {1} - x _ {1} = 3 x _ {1}$

The acceleration of a moving body is found from the slope of velocity-time graph.

$a = \frac {\mathrm{d} v}{\mathrm{d} t} = \text { Slope of } v t \text { curve }$

Time to reach the maximum height
$t _ {1} = u / g$
If $t_2$ be the time taken to hit the ground
$- H = u t _ {2} - \frac {1}{2} g t _ {2} ^ {2}$
But $t_2 = nt_1$ (given)
$\Rightarrow - H = u \frac {n u}{g} - \frac {1}{2} g \frac {n ^ {2} u ^ {2}}{g ^ {2}}$
$\Rightarrow 2 g H = n u ^ {2} (n - 2).$

The distance $ h_1 $ covers by stone,
$h _ {1} = \frac {1}{2} g t ^ {2} \times \frac {1}{2} 1 0 \times (5 t) ^ {2} = 1 2 5$
The distance $ h_2 $ covers by stone,
$h _ {2} = \frac {1}{2} \times 1 0 \times (1 0 t) ^ {2} - \frac {1}{2} \times 1 0 \times (5 t) ^ {2} = 3 7 5$
The distance $ h_3 $ covers by stone,
$h _ {3} = \frac {1}{2} \times 1 0 \times (1 5 t) ^ {2} - \frac {1}{2} \times 1 0 \times (1 0 t) ^ {2} = 6 2 5$
The relation between $ h_b $ $ h_2 $ and $ h_3 $ is $ h_t = \frac{h_2}{3} = \frac{h_3}{5} $

When velocity of a body changes by equal amount in equal intervals of time, then the body is said to have uniform acceleration, this holds true for straight line motion.

Velocity-time graph gives the instantaneous value of acceleration at any instant. For non-uniformly accelerated motion, $ \nu $ t graph is non-linear.

The initial velocity of the body, $ u^2 = 2gh $
$\Rightarrow \quad u = \sqrt {2 g h} = \sqrt {2 \times 1 0 \times 1 2 5} = \sqrt {2 5 0 0} = 5 0 \mathrm{ms} ^ {- 1}$
The final velocity from equation of motion,
$v ^ {2} = u ^ {2} + 2 g h$
$v = \sqrt {(5 0) ^ {2} + 2 \times 1 0 \times 5 5}$
$v = \sqrt {2 5 0 0 + 1 1 0 0}$
$= \sqrt {3 6 0 0} = 6 0 \mathrm{ms} ^ {- 1}$

Given, velocity of water, $ v_{w} = 4 \mathrm{~ms}^{-1} $
Velocity of person, $ v_{p} = 5 \mathrm{~ms}^{-1} $, width of the river $ = 846 \mathrm{~m} $
We know that, time taken, $ t = \frac{s}{\sqrt{v_p^2 - v_w^2}} $
$\Rightarrow \quad t = \frac {8 4 6}{\sqrt {2 5 - 1 6}}$
$\Rightarrow \quad t = \frac {8 4 6}{3} \cong 2 8 2 \mathrm{s}$

From graph, velocity of robber's car = 10 ms $ ^{-1} $
Let police car crosses it after t second.
Distance travelled by robber's car = 10 t m
Police car is moving with a constant acceleration of $ 1 \, ms^{-2} $ as it attains a velocity of $ 10 \, ms^{-1} $ in 10 s after starting from rest.
Distance travelled in t second = $ \frac{1}{2} $ a · $ t^{2} $ = $ \frac{1}{2} $ t $ ^{2} $
When the police car crosses the robber's car, distance
travelled by the both cars should be same from the starting point of chase.
$\therefore \quad \frac {1}{2} t ^ {2} \equiv 1 0 t \Rightarrow t = 2 0 s$

Given, initial speed, $ u = 1\mathrm{kms}^{-1} = 1000\mathrm{ms}^{-1} $
Final speed, $ v = 9 \, \text{kms}^{-1} = 9000 \, \text{ms}^{-1} $
By using the relation, $ v^{2} = u^{2} + 2as $
$(9 0 0 0) ^ {2} = (1 0 0 0) ^ {2} + 2 \times a \times 4 \Rightarrow a = 1 0 ^ {7} \mathrm{ms} ^ {- 2}$
∴ The time for which the particle remains in the tube
v = u + a t
$\Rightarrow \quad t = \frac {v - u}{a} = \frac {9 0 0 0 - 1 0 0 0}{1 0 ^ {7}} = 8 \times 1 0 ^ {- 4} \mathrm{s}$

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

Given, $x = 8 + 12t - t^3$
We know that, $ v = \frac{dx}{dt} $ and $ a = \frac{dv}{dt} $
$\mathrm{So}, \quad v = 1 2 - 3 t ^ {2}$
and $t = 2\mathrm{s}$
v = 0
and $a = -12\mathrm{ms}^{-2}$
So, retardation of the particle $= 12\mathrm{ms}^{-2}$.

Distance, $x = b_{0} + b_{1}t + b_{2}t^{2}$
Velocity, $ v = \left(\frac{dx}{dt}\right) = b_1 + 2b_2t $
Acceleration, $a = \frac{d^2x}{dt^2} = 2b_2$

Let $ t $ second be the time of flight of the first body after meeting, then $ (t - 4) $ second will be the time of flight of the second body.
As the initial velocity at which the bodies A and B projected are same and also the position of meeting will be also same.
So, $h_x = h_y$
$\therefore \quad 9 \partial t - \frac {1}{2} g t ^ {2} = 9 \partial (t - 4) - \frac {1}{2} g (t - 4) ^ {2}$
On solving, t = 12 s

Let after time t, car catches the scooter and the distance travelled by scooter in time t,
$x = \frac {1}{2} \times (1) \times t ^ {2} = \frac {t ^ {2}}{2} \quad \dots . (1)$
The distance travelled by car in time t
$x + 1 5 0 = \frac {1}{2} \times 2 \times t ^ {2} = t ^ {2} \quad \dots . \tag {11}$
Solving Eqs. (i) and (ii), we get
$t = \sqrt {3 0 0} \mathrm{s}$

Given, $\mathbf{r}_1(t) = 3t\hat{\mathbf{i}} +4t^2\hat{\mathbf{j}}$
$\therefore \quad \frac {d \mathbf {r} _ {1}}{d t} = 3 \hat {\mathbf {i}} + 6 \hat {\mathbf {j}}$
At $t = 1\mathrm{s},$
$\mathbf {v} _ {1} = \frac {d \mathbf {r} _ {1}}{d t} = 3 \hat {\mathbf {i}} + 6 \hat {\mathbf {i}}$
Again, $\mathbf{r}_2(0) = 4t^2\hat{\mathbf{i}} +3\hat{\mathbf{j}}$
$\Rightarrow \quad \frac {d \mathbf {r} _ {2}}{d t} = 8 \hat {\mathbf {i}} + 3 \hat {\mathbf {j}}$
At 1=1s,
$\mathbf {v} _ {2} = \frac {d \mathbf {r} _ {2}}{d t} = 8 \hat {\mathbf {i}} + 3 \hat {\mathbf {J}}$
Relative speed, $\mathbf{v}_{\mathrm{rel}} = \mathbf{v}_2 - \mathbf{v}_1$
$= (8 \hat {\mathbf {i}} + 3 \hat {\mathbf {j}}) - (3 \hat {\mathbf {i}} + 8 \hat {\mathbf {j}})$
$= 5 \hat {\mathbf {i}} - 5 \hat {\mathbf {j}}$
$\therefore \quad | \mathbf {v} _ {\mathrm{rel}} | = \sqrt {(5) ^ {2} + (- 5) ^ {2}}$
$= 5 \sqrt {2} \mathrm{m/s}$

D

Time taken by the ball to reach highest point is $t = 1s$ . As the person throws the second ball, when the velocity of the first ball becomes zero, i.e., $v = 0$ or when the first ball reaches the highest point. Using, $v = u + at$ Here, $v = 0$ , $a = -g$ , $t = 1s$ $\therefore 0 = u - (10)(1) \Rightarrow u = 10m/s$ Using $v^2 - u^2 = 2ah$ , we get $(0)^2 - (10)^2 = 2(-10)(h) \Rightarrow h = \frac{(10)^2}{20} = 5m$ .

Given: $x = 3t^3$ ; $y = 4t^3$ $\therefore v_x = \frac{dx}{dt} = \frac{d}{dt}(3t^3) = 9t^2$ $v_y = \frac{dy}{dt} = \frac{d}{dt}(4t^3) = 12t^2$ The velocity of the body is $v = \sqrt{v_x^2 + v_y^2} = \sqrt{(9t^2)^2 + (12t^2)^2} \Rightarrow v = 15t^2$ .

$x = 8 + 12t - t^3$ $v = 0 + 12 - 3t^2 = 0$ $3t^2 = 12$ $t = 2\sec$ $a = \frac{dv}{dt} = 0 - 6t$ $a[t = 2] = -12m / s^2$ retardation $= 12m / s^2$ .

C

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=50 s $S_{BA}=u_{BA}t+\frac{1}{2}a_{BA}t^{2}=0\times60+\frac{1}{2}\times1\times(50)^{2}=1250m.$