Kinematics-1D

Given, $x = (t - 2)^{2}$
Velocity, $v = \frac{dx}{dt} = \frac{d}{dt} (t - 2)^2 = 2(t - 2)\mathrm{ms}^{-1}$
Acceleration, $a = \frac{dv}{dt} = \frac{d}{dt} (2t - 2) = 2(1 - 0) = 2\mathrm{ms}^{-2}$
When $ t = 0 $; $ v = -4 \, \mathrm{ms}^{-1} $, $ t = 2 \, \mathrm{s} $; $ v = 0 \, \mathrm{ms}^{-1} $
$t = 4 \mathrm{s}; v = 4 \mathrm{ms} ^ {- 1}$
v-t graph is shown in figure below.

Distance travelled = Area under the v-t graph
$= | \text { Area OAC } | + \text { Area ABD }$
$= \left| \frac {4 \times 2}{2} \right| + \frac {1}{2} \times 2 \times 4 = 8 \mathrm{m}$

Let height of tower is h and body takes t time to reach to ground when it fall freely.
$\therefore \quad h = \frac {1}{2} g t ^ {2} \quad \dots \tag {i}$
In last second, i.e. tth second body travels = 0.36h
It means in rest of the time, i.e. in $ (t-0) $ second, it travels
= h - 0. 3 6 h = 0. 6 4 h
Now, applying equation of motion for $ (t-0) $ second

Acceleration, $ f = f_{0}\left(1 - \frac{t}{T}\right) $
$\frac {d v}{d t} = f _ {0} \cdot \left(1 - \frac {t}{T}\right) \quad \left(\because f = \frac {d v}{d t}\right)$
$\Rightarrow \quad d v = f _ {0} \cdot \left(1 - \frac {t}{T}\right) d t \Rightarrow \int d v = \int f _ {0} \left(1 - \frac {t}{T}\right) d t$
$v = f _ {0} t - f _ {0} \frac {t ^ {2}}{2 T} + C \quad \dots \tag {1}$
where, C is constant.
When $ t = 0, v = 0 $ thus from Eq. (i), $ C = 0 $
$v = f _ {0} t - \frac {f _ {0}}{T} \cdot \frac {t ^ {2}}{2} \quad \dots \tag {ii}$
As, $f = f_0\left(1 - \frac{t}{T}\right)$
When $f = 0$
$\therefore \quad f _ {0} \left(1 - \frac {t}{T}\right) = 0; f _ {0} \neq 0 \Rightarrow t = T$
Putting $ t = T $ in Eq. (ii), we get
$v = f _ {0} T - \frac {f _ {0}}{T} \cdot \frac {T ^ {2}}{2} = \frac {f _ {0} T}{2}$

Relative to lift,
$u _ {s} = 0, \alpha_ {s} = (9. 8 + 1. 2) = 1 1 \mathrm{ms} ^ {- 2}$
$s _ {s} = \frac {1}{2} \alpha_ {s} t ^ {2} \Rightarrow 2. 7 = \frac {1}{2} \times 1 1 \times t ^ {2} \Rightarrow t = \sqrt {\frac {5 4}{1 1}} \mathrm{s}$

After bailing out from point A, parachutist falls freely under gravity. The velocity acquired by it will be v.
$\text { From } v ^ {2} = u ^ {2} + 2 a s = 0 + 2 \times 9. 8 \times 5 0 = 9 8 0$
$(\mathrm{As}, u = 0, a = 9. 8 \mathrm{ms} ^ {- 2}, s = 5 0 \mathrm{m})$
At point B, parachute opens and it moves with retardation of $ 2 \, ms^{-2} $ and reach at ground (point C with velocity of $ 3 \, ms^{-1} $ ).
For the part BC by applying the equation $ v^{2}=u^{2}+2as $
$v = 3 \mathrm{ms} ^ {- 1}, u = \sqrt {9 6 0} \mathrm{ms} ^ {- 1}, a = - 2 \mathrm{ms} ^ {- 2}, s = h$
$\Rightarrow \quad (3) ^ {2} = (\sqrt {9 6 0}) ^ {2} + 2 \times (- 2) \times h$
$\Rightarrow \quad 9 = 9 8 0 - 4 h$
$\Rightarrow \quad h = \frac {9 8 0 - 9}{4} = \frac {9 7 1}{4} = 2 4 2. 7 \equiv 2 4 3 \mathrm{m}$
So, the total height by which parachutist bails out
$= 5 0 + 2 4 3 = 2 9 3 \mathrm{m}$

Initial relative velocity $= v_{1} - v_{2}$, final relative velocity $= 0$
$\text { From } \quad v ^ {2} = u ^ {2} - 2 a s \Rightarrow 0 = \psi_ {1} - v _ {2}) ^ {2} - 2 \times a \times s$
$\Rightarrow \quad s = \frac {\psi_ {1} - v _ {2}) ^ {2}}{2 a}$
If the distance between two cars is s, then collision will take place. To avoid collision d > s
$\therefore \quad d > \frac {\psi_ {1} - v _ {2}) ^ {2}}{2 a}$
where, d = actual initial distance between two cars.

Time taken by first drop to reach the ground, $ t = \sqrt{\frac{2h}{g}} $
$\Rightarrow \quad t = \sqrt {\frac {2 \times 5}{1 0}} = 1 \mathrm{s}$
As the water drops fall at regular intervals from a tap,
therefore time difference between any two drops $ = \frac{1}{2} $ s.
In this given time, distance of second drop from the tap
$= \frac {1}{2} g \left(\frac {1}{2}\right) ^ {2} = \frac {1 0}{8} = 1. 2 5 \mathrm{m}$
Its distance from the ground = 5 - 1.25 = 3.75 m

$A \Rightarrow \frac{u^2}{4} - u^2 = -2gh_0$
$B \Rightarrow \frac {u ^ {2}}{9} - u ^ {2} = - 2 g h _ {2}$
$C \Rightarrow \frac {u ^ {2}}{1 6} - u ^ {2} = - 2 g h _ {3}$

$\therefore AB = \frac{u^2}{2g}\left(\frac{8}{9} -\frac{3}{4}\right) = \frac{u^2}{2g} -\frac{5}{36}$
$B C = \frac {u ^ {2}}{2 g} \left(\frac {1 5}{1 6} - \frac {8}{9}\right) = \frac {u ^ {2}}{2 g} - \frac {7}{1 4 4}$
$\therefore \quad \frac {A B}{B C} = \frac {5}{3 6} \times \frac {1 4 4}{3 6} = \frac {2 0}{7}$

For the given condition, initial height $ h = d $ and velocity of the ball is zero. When the ball moves downward, its velocity increases and it will be maximum, when the ball hits the ground and just after the collision, it becomes half and in opposite direction. As the ball moves upwards, its velocity again decreases and becomes zero at height $ d / 2 $. This explanation match with graph (a).

The position $ x $ of a particle w.r.t. time $ t $ along $ X $-axis, $ x = 9x^{2} - t^{3} $ ... (i)
Differentiating Eq. (i) w.r.t. time, we get speed, i.e.
$v = \frac {d x}{d t} = \frac {d}{d t} (9 x ^ {2} - t ^ {3})$
$\Rightarrow \quad v = 1 8 t - 3 t ^ {2} \tag {ii}$
Again, differentiating Eq. (ii) w.r.t. time, we get acceleration, i.e.
$a = \frac {d v}{d t} = \frac {d}{d t} (1 8 t - 3 t ^ {2})$
$\Rightarrow \quad a = 1 8 - 6 t \tag {iii}$
Now, when speed of particle is maximum, its acceleration is zero.
a = 0
$\Rightarrow \quad 1 8 - 6 t = 0$
$\Rightarrow \quad t = 3 \mathrm{s}$
Putting in Eq. (i), we obtain position of particle at that time
$x = 9 (3) ^ {2} - (3) ^ {3} = 9 (9) - 2 7$
$= 8 1 - 2 7 = 5 4 \mathrm{m}$

Relative velocity of car A w.r.t. ground
$v _ {A y} = 3 0 \mathrm{km/h}$
$\mathbf{v}_{Ay} = 40\mathrm{km / h}$
∴ Relative velocity of A w.r.t. B,
$\mathbf {v} _ {A B} = \mathbf {v} _ {A y} - \mathbf {v} _ {B y} = \mathbf {v} _ {A y} + \left(- \mathbf {v} _ {B y}\right)$
$\left| \mathbf {v} _ {A B} \right| = \sqrt {(3 0) ^ {2} + (4 0) ^ {2}} = 5 0 \mathrm{km/h}$
$\therefore \quad \tan \theta = \frac {3 0}{4 0} = \frac {3}{4}$
$\Rightarrow \quad \theta = \tan^ {- 1} (3 / 4) \mathrm{N} \text { of west }$

$\left| \begin{array}{c} \text { Relative velocity of man w.r.t. ground } \\ \mathbf {v} _ {m y} > 3 \mathrm{m/s} \end{array} \right.$
Relative velocity of rain w.r.t. ground,
$\mathbf {v} _ {y y} = 4 \mathrm{m/s}$
Relative velocity of rain w.r.t. man, $ v_{m} = ? $
$\mathbf {v} _ {r m} = \mathbf {v} _ {r g} - \mathbf {v} _ {m g} = \mathbf {v} _ {r g} + (- \mathbf {v} _ {m g})$

$\mathbf {v} _ {r m} = 5 \mathrm{m/s}$
$\tan \alpha = \frac {3}{4}, \tan \beta = \frac {4}{3}$
$\mathbf{v}_{rm} = 5\mathrm{m / s}$ at $\tan^{-1}\left(\frac{3}{4}\right)$ with vertical or at $\tan^{-1}\left(\frac{4}{3}\right)$ with horizontal.

where, $\mathbf{v}_{sg}$ is the velocity of boat w.r.t. ground and $\mathbf{v}_{sg}$ is the velocity of ship w.r.t. ground.
Relative velocity of boat w.r.t. ship $\mathbf{v}_{sg}$ is along north.
$\mathbf{v}_{bs}$ is resultant of $\mathbf{v}_{sg}$ and $\mathbf{v}_{sg}$ in opposite direction.
$v \cos 3 0 ^ {\circ} = v _ {b s}$
$v \sin 3 0 ^ {\circ} = 1 5 \Rightarrow v = 3 0 \mathrm{km/h}$

$u: v_{sg}$ (velocity of river w.r.t. ground)
$v: v_{ms}$ (velocity of man w.r.t. river)
Along the flow, $ v_{mg} = v + u $, $ 3 = \frac{d}{v + u} $ (m)
where, $ v_{mg} $ is velocity of man w.r.t. ground.
Opposite to the flow,
$v _ {m g} = v - u, 6 = \frac {d}{v - u} \tag {iii}$
$\Rightarrow \quad t _ {0} = \frac {d}{v} = 4 \mathrm{h}$

y: displacement = velocity × time

When a particle is released from rest, $ v = 0 $ but $ a \neq 0 $.

$a = \frac{\mathrm{d}v}{\mathrm{d}t} =$ slope of $v - t$ graph.
Perpendicular to t-axis, slope = ∞, therefore a = ∞

Slope of $ s - t $ graph = Velocity = Positive
At $ t = 0 $, $ s \neq 0 $. Further at $ t = t_0 $; $ s = 0 $, $ v \neq 0 $.
Thus, v is constant.

By differentiating $a - t$ equation two times, we will get $s - t$ equation.
Further

Ascending means velocity is in upward direction. Speed is decreasing. It means acceleration is downwards.
Further, the body moves in the direction of velocity.

From the given x-t graph, it is clear that
at $t = \Omega ,\frac{dx}{dt} = 0$
∴ $ V_{velocity}, v = \frac{dx}{dt} = \Omega $ at t = 0
As time passes $\frac{dx}{dt}$ increases and then decreases.
∴ Velocity and hence, acceleration changes.

$x = 4(t - 2) + a(t - 2)^{2}$
At $t = \Omega, x = -8 + 4a = 4a - 8$
$v = \frac {d x}{d t} = 4 + 2 a (t - 2)$
At $t = 0, v = 4 - 4a = 4(1 - a)$
But acceleration, $a = \frac{d^2x}{dt^2} = 2a$

For maximum and minimum displacement, we have to keep in mind the magnitude and direction of maximum velocity.
As maximum velocity in positive direction is $ v_{0} $ maximum velocity in opposite direction is also $ v_{0} $ .
Maximum displacement in one direction $= v_{0}T$
Maximum displacement in opposite directions $= -v_{0}T$
Hence, $-v_{0}T < x < v_{0}T$.

Distance travelled in last second of upward journey is independent of velocity of projection.
Distance travelled $= \frac{1}{2} g t^2 = \frac{1}{2}\times 9.8\times (10^{2} = 4.9\mathrm{m}$

Average velocity $= \frac{\text{Displacement}}{\text{Time}}$
$= \frac {\text { Area of } v - t \text { graph }}{\text { Time }}$
$= \frac {(1 / 2) v _ {m} t _ {0}}{t _ {0}} = \frac {1}{2} v _ {m}$