Kinematics-1D

$(0,0)\xrightarrow{A}\frac{x_1}{A}\xrightarrow{C}\frac{x_2}{A}\xrightarrow{B}$ $u = 0\longleftarrow \longrightarrow r2\longrightarrow \longrightarrow r2\longrightarrow$
$a = \text{constant}$
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.

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

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
$\Rightarrow \quad t = \frac {v - u}{a} = \frac {9 0 0 0 - 1 0 0 0}{1 0 ^ {7}} = 8 \times 1 0 ^ {- 4} \mathrm{s}$

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, $\mathbf{v}_{\mathrm{rel}} = \mathbf{v}_2 - \mathbf{v}_1$
$= (8 \hat {\mathbf {i}} + 3 \hat {\mathbf {j}}) - (3 \hat {\mathbf {i}} + 8 \hat {\mathbf {j}})$
$= 5 \hat {\mathbf {i}} - 5 \hat {\mathbf {j}}$
$\therefore \quad | \mathbf {v} _ {\mathrm{rel}} | = \sqrt {(5) ^ {2} + (- 5) ^ {2}}$
$= 5 \sqrt {2} \mathrm{m/s}$

Average speed of a boy $ =\frac{2v_{1}v_{2}}{v_{1}+v_{2}}=\frac{2\times2.5\times4}{2.5+4} $
$= \frac {2 0}{6 . 5} = \frac {4 0}{1 3} \mathrm{km/h}$

$= \frac {1}{2} g t ^ {2} \text {or} t = \sqrt {\frac {4}{g}}$
But $t_2 = t - t_1 = \sqrt{\frac{4}{g}} -\sqrt{\frac{2}{g}}$
$\therefore \quad \frac {t _ {1}}{t _ {2}} = \frac {\sqrt {2 / g}}{\sqrt {4 / g} - \sqrt {2 / g}} = \frac {\sqrt {2}}{2 - \sqrt {2}}$
$= \frac {\sqrt {2} (2 + \sqrt {2})}{2} = (\sqrt {2} + 1)$

Time taken to travel first half distance, $ t_{\mathrm{f}} = \frac{1 / 2}{v_1} = \frac{L}{2v_1} $
Time taken to travel second half distance, $ t_2 = \frac{L}{2v_2} $
$\text { Total time } = t _ {\mathrm{f}} + t _ {2} = \frac {L}{2 v _ {1}} + \frac {L}{2 v _ {2}} = \frac {L}{2} \left(\frac {1}{v _ {1}} + \frac {1}{v _ {2}}\right)$
We know that, $ v_{\mathrm{av}} = $ Average speed
$= \frac {\text { total distance }}{\text { total time }} = \frac {L}{\frac {L}{2} \left(\frac {1}{v _ {1}} + \frac {1}{v _ {2}}\right)} = \frac {2 v _ {1} v _ {2}}{v _ {1} + v _ {2}}$

In all the given graphs, we have more than one value of speed corresponding to a single time. However, the graph must have a unique value of speed corresponding to a single time. Thus, all the three graphs are not possible.

0 0 = 1 0 t
$\Rightarrow \quad t = 5 0 \mathrm{s}$

Man walks from his home to market with a speed of 5 kmh $ ^{-1} $ , Distance = 2.5 km and time
$= \frac {d}{v} = \frac {2 . 5}{5} = \frac {1}{2} = 3 0 \mathrm{min}$
and he returns back with speed of 7.5 kmh $ ^{-1} $ in rest a time of 20 min.
Distance = $ 7.5 \times \frac{20}{60} = 2.5 $ km
So, average speed = $ \frac{\text{total distance}}{\text{total time}} = \frac{(2.5 + 2.5)}{(50 / 60) \text{ h}} = \frac{30}{5} \text{ km/h} $

$ s \propto u^{2} $ . If u becomes 3 times, then s will become 9 times, i.e. $ 9 \times 20 = 180 \, m $

Distance travelled by body A in 5th second and distance travelled by body B in 3rd second of its motion are equal.

: Slope of displacement-time graph at the point E is negative. Thus, at the point E, the intantaneous velocity of the particle is negative.

Since $ x = 1.2t^2 $, comparing it with equation of motion
$x = \frac {1}{2} a t ^ {2}.$
Thus, the motion is uniformly accelerated.

It is given that, $ v = \frac{t^2}{10} + 20 $
Differentiating both sides w.r.t. time, we get
$\frac {d v}{d t} = \frac {2 t}{1 0} + 0 = \frac {t}{5}$
$\therefore$ Acceleration, $a = \frac{dv}{dt} = \frac{t}{5} x$ constant
Hence, it is a case of non-uniform acceleration.