Kinematics-1D

Motion is first uniformly accelerated in positive direction, then it is uniform in negative direction.

$h = -3t + 5t^2$ and $2l = 3t + 5t^2$
$\begin{array}{c} P \\ 2 h \end{array} \left. \begin{array}{c} \uparrow \text { s m s } ^ {- 1} \\ \uparrow \text { s m s } ^ {- 1} \\ \uparrow \text { s m s } ^ {- \frac {1}{2}} \\ \uparrow \text { s m s } ^ {- \frac {1}{2}} \\ \uparrow \text { s m } \end{array} \right\}$
From these two equations, we get $ t = 3 $ s and $ h = $ distance between $ P $ and $ Q = 30 $ m.

Let height of Minaret be $H$ and body take time $T$ to fall from top to bottom.

In last $2\text{ s}$, body travels distance of $40\text{ m}$, so in $(T-2)$ second, distance travelled $=(H-40)\text{ m}$
$(H-40)=\frac{1}{2}g(T-2)^2$
$\qquad \text{...(ii)}$
By solving Eqs. (i) and (ii), we get
$T=3\text{ s} and H=45\text{ m}$

Given acceleration-time graph is shown in figure.

Let $a(t=0)=a_0=\text{constant}$
From the graph, we have
$\frac{t}{T}+\frac{a}{a_0}=1$
$\Rightarrow a_0t+aT=a_0T\qquad \text{...(i)}$
Also, we know that,
$a=\frac{dv}{dt}\qquad \text{...(ii)}$
From Eqs. (i) and (ii), we get
$\frac{a_0T-a_0t}{T}=\frac{dv}{dt}$
Integrating both sides, we get
$\int_ {0} ^ {t} \left(\mu_ {0} T - \alpha_ {0} \mu\right) d t = T \int_ {0} ^ {t} d v$
$\Rightarrow \quad \alpha_ {0} T t - \frac {\alpha_ {0} t ^ {2}}{2} = T v \Rightarrow v = \alpha_ {0} t - \frac {\alpha_ {0}}{2 T} t ^ {2}$
Therefore, v-t graph must be parabolic.
Now, at $t = 0$ $v = 0$
$t = T _ {v} v = \frac {\alpha_ {0} T}{2} = \text { constant }$
These conditions are satisfied by graph (a).

Given line has positive intercept but negative slope. So, its equation can be written as
$v = - m x + v _ {0} \left(\text { where }, m = \tan \theta = \frac {v _ {0}}{x _ {0}}\right) \quad \dots \tag {i}$
By differentiating with respect to time, we get
$\frac {d v}{d t} = - m \frac {d x}{d t} = - m v$
Now, substituting the value of $ v $ from Eq. (i), we get
$\frac {d v}{d t} = - m (- m x + v _ {0}) = m ^ {2} x - m v _ {0}$
$\therefore \quad a = m ^ {2} x - m v _ {0}$
i.e. The graph between $a$ and $x$ should have positive slope but negative intercept on $a$-axis. So, graph (a) is correct.

$\frac{dx}{dt} = v = -1 - 2t$
Comparing with $ v = u + \alpha t $, we get
$u = - 1 \mathrm{ms} ^ {- 1} \text { and } \alpha = - 2 \mathrm{ms} ^ {- 2}.$
At $ t = 0 $, $ x = 1\mathrm{m} $. Then, $ u $ and $ \alpha $ both are negative. Hence,
x-coordinate of particle will go on decreasing.

In graph (b) for one value of displacement, there are two different points of time. Hence, for one time, the average velocity is positive and for other time is equivalent negative.
As there are opposite velocities in the interval 0 to $ T_{1} $, hence average
velocity can vanish in (b). This can be seen in the figure alongside.
Here, $OA = BT$ (same displacement) for two different points of time.

As the lift is coming in downward directions, displacement will be negative. We have to see whether the motion is accelerating or retarding.
We know that due to downward motion, displacement will be negative. When the lift reaches 4th floor and is about to stop, hence motion is
retarding in nature, hence $x < \mathcal{O}_1 > 0$.
As displacement is in negative direction, velocity will also be negative, i.e. $ v < 0 $.

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 = ∞