Kinematics-1D

Taking downward direction as positive CASE-1
$h = u t _ {1} + \frac {1}{2} g t _ {1} ^ {2}\tag{... (1}$
CASE-2
$h = - u t _ {2} + \frac {1}{2} g t _ {2} ^ {2}\tag{... (2}$
$(1) t _ {2} + (2) t _ {1} \mathrm{gives}$
$h \left(t _ {1} + t _ {2}\right) = \frac {1}{2} g t _ {1} ^ {2} t _ {2} + \frac {1}{2} g t _ {2} ^ {2} t _ {1}$
$\Rightarrow \quad h = \frac {1}{2} g t _ {1} t _ {2}\tag{... (3}$
If $t$ is the time taken by it to fall freely to the ground from the same height, then
$h = \frac {1}{2} g t ^ {2}\tag{... (4}$
Equating (3) and (4), we get
$t = \sqrt {t _ {1} t _ {2}}$
Hence, the correct answer is (D).

$a = k v\tag{87.}$
$v \frac {d v}{d x} = k v$
$\left\{\because a = \frac {d v}{d t} = v \frac {d v}{d x} \right\}$
$\Rightarrow \frac {d v}{d x} = k$
$\Rightarrow$ Slope of velocity-displacement graph is a constant Hence, the correct answer is (C).

Velocity of 1st particle at time t is $v_{1}=v$
Velocity of 2nd particle at time t is $v_{2} = at$
Relative velocity $v_{r}=\sqrt{v_{1}^{2}+v_{2}^{2}-2v_{1}v_{2}\cos\alpha}$
$\Rightarrow v _ {r} ^ {2} = v ^ {2} + a ^ {2} t ^ {2} - 2 v (a t) \cos \alpha$
$\begin{array}{c}\mathbf {v} _ {2} = a t\\\alpha\end{array}\rightarrow v _ {1} = v$
For $v_{r}$ to be least or $v_{r}^{2}$ to be least we must have
$\frac {d}{d t} \Big (v _ {r} ^ {2} \Big) = 0$
$\Rightarrow 0 + 2 a ^ {2} t - 2 a v \cos \alpha = 0$
$\Rightarrow t = \frac {v \cos \alpha}{a}$
Hence, the correct answer is (B).

$v _ {r} = \sqrt {v ^ {2} + a ^ {2} \left(\frac {v ^ {2} \cos^ {2} \alpha}{a ^ {2}}\right) - 2 v \left(\frac {v \cos \alpha}{a}\right) a \cos \alpha}$
$\Rightarrow v _ {r} = \sqrt {v ^ {2} - v ^ {2} \cos^ {2} \alpha}$
$\Rightarrow v _ {r} = v \sin \alpha$
Hence, the correct answer is (A).

$v_{A}^{2}=2g(h+x)$ and
$\begin{array}{r l} & v _ {B} ^ {2} = u ^ {2} + 2 g x \\ \text {Since} & v _ {A} = v _ {B} \\ \Rightarrow & u ^ {2} + 2 g x = 2 g h + 2 g x \\ \Rightarrow & u ^ {2} = 2 g h \\ \Rightarrow & u = \sqrt {2 g h} \end{array}$
Hence, the correct answer is (C).

$v_{1} + v_{2} = 4$
$\begin{array}{r l} & v _ {1} - v _ {2} = \frac {4}{1 0} = 0. 4 \\ \Rightarrow & 2 v _ {1} = 4. 4 \\ \Rightarrow & v _ {1} = 2. 2 \mathrm{ms} ^ {- 1} \text { and } v _ {2} = 1. 8 \mathrm{ms} ^ {- 1} \\ & \text { Hence,the correct answer is(A). } \end{array}$

$\begin{array}{r l} & {\frac {g}{2} (2 n - 1) = \frac {1}{2} g (3) ^ {2}} \\ & {\Rightarrow 2 n = 1 0} \\ & {\Rightarrow n = 5 \mathrm{s}} \end{array}$
Hence, the correct answer is (A).

$t = \frac {a}{v - v \cos \left(\frac {2 \pi}{n}\right)} = \frac {a}{2 v \sin^ {2} \left(\frac {\pi}{n}\right)}$

where n is the number of sides in a symmetrical polygon. For this problem n = 6
$\Rightarrow \quad t = \frac {a}{2 v \sin^ {2} \left(\frac {\pi}{6}\right)} = \frac {2 a}{v}$
Hence, the correct answer is (A).

METHOD I
$v = k x + v _ {0}$
$\mathrm{Since} a = \frac {d v}{d t}$
$\Rightarrow a = k \left(\frac {d x}{d t}\right)$
$\Rightarrow a = k v = k ^ {2} x + k v _ {0}$
$\Rightarrow a = \alpha x + \mathrm{constant}$
METHOD II
$\mathrm{Since} v = k x + v _ {0}$
$\Rightarrow \frac {d v}{d x} = k$
$\mathrm{Now} a = \frac {d v}{d t} = v \left(\frac {d v}{d x}\right)$
$\Rightarrow a = \left(k x + v _ {0}\right) k$
$\Rightarrow a = k ^ {2} x + k v _ {0}$
$\Rightarrow \quad a = \alpha x + \text { constant }$
So, we observe that acceleration increases linearly with x.
Hence, the correct answer is (D).

Area under a-t graph equals the change in velocity ( $\Delta v$ )
$\begin{array}{r l} \Rightarrow & \Delta v = \frac {1}{2} (3) (4) = 6 \mathrm{ms} ^ {- 1} \\ \Rightarrow & v _ {f} - v _ {i} = 6 \\ \Rightarrow & v _ {f} - 2 = 6 \\ \Rightarrow & v _ {f} = 8 \mathrm{ms} ^ {- 1} \end{array}$
Hence, the correct answer is (D).

From ground to maximum height, we have
$\begin{array}{r l} & h = \frac {u ^ {2}}{2 g} = \frac {(2 0) ^ {2}}{2 (9 . 8)} \\ \Rightarrow & h = \frac {4 0 0}{1 9 . 6} = 2 0. 4 \mathrm{m} \\ \Rightarrow & t _ {1} = \frac {v}{g} = \frac {2 0 . 4}{9 . 8} = 2. 0 8 \mathrm{s} \end{array}$
Now the ball is caught 5 m above the point of launch in its downward motion. So, it is travelled 15 m down in time $t_{2}$ (say).
$\mathrm{So}, 1 5 = 0 + \frac {1}{2} g t _ {2} ^ {2}$
$\Rightarrow 1 5 = 4. 9 t _ {2} ^ {2}$
$\Rightarrow t _ {2} = \sqrt {3 . 0 6}$
$\Rightarrow t _ {2} = 1. 7 5 \mathrm{s}$
So, total time $T = t_{1} + t_{2}$
$\Rightarrow T = 2. 0 8 + 1. 7 5$
$\Rightarrow T = 3. 8 3 \mathrm{s}$
Please observe that if the options given are not close to each other, then you can also take $g=10\ ms^{-2}$ and the answer will be 3.73 s, which is very close to 3.8 s.
DO NOT APPLY THIS WHEN ANSWERS ARE VERY CLOSE IN VALUES.
Hence, the correct answer is (B).

At time t, the separation between the particles is
$x = x _ {1} - x _ {2} = 2 v t - \frac {1}{2} a t ^ {2}$
For $x$ to be MAXIMUM,
${\frac {d x}{d t}} = 0$
2 v - a t = 0
$\Rightarrow t = \frac {2 v}{a}$
$x _ {M A X} = (2 v) \left(\frac {2 v}{a}\right) - \frac {1}{2} a \left(\frac {4 v ^ {2}}{a ^ {2}}\right)$
$\Rightarrow x _ {M A X} = \frac {4 v ^ {2}}{a} - \frac {2 v ^ {2}}{a} = \frac {2 v ^ {2}}{a}$
Hence, the correct answer is (C).

Since, area under a-t graph is equal to the change in velocity, so
$\Delta v = \frac {1}{2} (2 + 4) (2) + \frac {1}{2} (2 + 4) (2) = 1 2 \mathrm{ms} ^ {- 1}$

$\Rightarrow \Delta v = v - u = 1 2 \mathrm{ms} ^ {- 1}$
Since $u = 0$
$\Rightarrow v = 1 2 \mathrm{ms} ^ {- 1}$
Hence, the correct answer is (B).

$v _ {a v} = \frac {\text { Total Distance Travelled }}{\text { Total Time Taken }}$
$\Rightarrow v _ {a v} = \frac {\frac {l}{3} + \frac {2 l}{3}}{t _ {1} + t _ {2}}\tag{... (1}$
$\begin{array}{c c c c c} & \frac {I}{3} & C & \frac {2 I}{3} & D \\ A & & & t _ {2} / 2 & t _ {2} / 2 \\ & t _ {1} & & t _ {2} \end{array}$
$\begin{array}{l} \text {For AC,} \frac {l}{3} = 4 t _ {1} \\ \Rightarrow t _ {1} = \frac {l}{1 2} \\ \text {For CB,} \frac {2 l}{3} = 2 \left(\frac {t _ {2}}{2}\right) + 6 \left(\frac {t _ {2}}{2}\right) \\ \Rightarrow \frac {2 l}{3} = t _ {2} + 3 t _ {2} \\ \Rightarrow t _ {2} = \frac {l}{6} \end{array}\tag{... (2}$
... (3)
Substituting (2) and (3) in (1), we get
$v _ {a v} = \frac {l}{\frac {l}{1 2} + \frac {2 l}{1 2}} = \frac {l}{\left(\frac {3 l}{1 2}\right)}$
$\Rightarrow v _ {a v} = 4 \mathrm{ms} ^ {- 1}$
Hence, the correct answer is (B).

Let velocities at A, B, C and D be $v_{A}$ , $v_{B}$ , $v_{C}$ and $v_{D}$ .

Since, $\ell = (v_{av})t$
For AB, $12 = \frac{v_A + v_B}{2} \Rightarrow v_A + v_B = 24$
... (1)
$\mathrm{ForCD,} 2 0 = \frac {v _ {C} + v _ {D}}{2} \Rightarrow v _ {C} + v _ {D} = 4 0\tag{... (2}$
For AB, $v_{B}^{2}-v_{A}^{2}=2al$
... (3)
For CD, $v_{D}^{2} - v_{C}^{2} = 2al$
$\Rightarrow \quad (v_B - v_A) = 24 \quad \text{and} \quad (v_D - v_C) = 40 \tag{4}$
$\Rightarrow \frac{v_B - v_A}{v_D - v_C} = \frac{5}{3}$
$\text{Let us assume:}$
$v_B - v_A = 5k \tag{5}$
$v_D - v_C = 3k \tag{6}$
$\text{Solving (1) and (5), we get:}$
$v_B = \frac{24 + 5k}{2}, \quad v_A = \frac{24 - 5k}{2} \tag{7}$
$\text{Solving (2) and (6), we get:}$
$v_D = \frac{40 + 3k}{2}, \quad v_C = \frac{40 - 3k}{2} \tag{8}$
$\text{Further, for interval } BC, \quad v_C^2 - v_B^2 = 2al \tag{9}$
$\text{Equating (9) and (3), and using (7) and (8), we get a quadratic in } k \text{:}$
$k^2 + 60k - 64 = 0$
$\text{We get } k = -30 + 2\sqrt{241} \quad (\text{Rejecting negative sign})$
$\therefore \text{Average velocity in interval } BC \text{ is:}$
$v_{\text{av}} = \frac{v_B + v_C}{2} = \frac{24 + 40 + 2k}{4}$
$\Rightarrow v_{\text{av}} = \left(1 + \sqrt{241}\right) \text{ ms}^{-1}$

Let $a_{1}$ and $a_{2}$ be the retardations offered to the bullet by wood and iron respectively.

For $A\to B\to C$
$v _ {1} ^ {2} - u ^ {2} = 2 a _ {1} (4), \mathrm{and}$
$0 ^ {2} - v _ {1} ^ {2} = 2 a _ {2} (1)$
Adding, we get
$- u ^ {2} = 2 \big (4 a _ {1} + a _ {2} \big)\tag{... (1}$
For $A^{\prime}\to B^{\prime}\to C^{\prime}$
$v _ {2} ^ {2} - u ^ {2} = 2 a _ {2} (2), \mathrm{and}$
$0 ^ {2} - v _ {2} ^ {2} = 2 a _ {1} (2)$
Adding, we get
$- u ^ {2} = 2 \left(2 a _ {1} + 2 a _ {2}\right)\tag{... (2}$
Equating (1) and (2) and solving, we get
$4 a _ {1} + a _ {2} = 2 a _ {1} + 2 a _ {2}$
$\Rightarrow a _ {2} = 2 a _ {1}$
Hence, the correct answer is (B).

$v_{x}=30\ ms^{-1}$
$\mathrm{Since}, h = \frac {1}{2} g t ^ {2}$
$\Rightarrow 8 0 = \frac {1}{2} (1 0) t ^ {2}$
$\Rightarrow t = 4 \mathrm{s}$
i.e., the ball hits the ground 4 s after the launch.
Now, since $v_{y}=gt=(10)(4)=40\ \mathrm{ms}^{-1}$
$\begin{array}{r l} \Rightarrow & v = \sqrt {v _ {x} ^ {2} + v _ {y} ^ {2}} \\ \Rightarrow & v = \sqrt {(3 0) ^ {2} + (4 0) ^ {2}} \\ \Rightarrow & v = 5 0 \mathrm{ms} ^ {- 1} \end{array}$
Hence, the correct answer is (B).

$v _ {1} = - \sqrt {2 g d} \mathrm{and} v _ {2} = + \sqrt {2 g \left(\frac {d}{2}\right)}$
Hence, the correct answer is (A).

Distance covered in the last one second i.e., distance covered in the nth second if the ball takes n seconds to hit the ground is
$x _ {1} = s _ {n ^ {t h}} = 0 + \frac {g}{2} (2 n - 1)\tag{... (1}$
Distance covered in the first three seconds is
$x _ {2} = \frac {1}{2} g (3) ^ {2} = \frac {9 g}{2}\tag{... (2}$
Since $x_{1}=x_{2}$
$\Rightarrow \frac {g}{2} (2 n - 1) = \frac {9 g}{2}$
$\Rightarrow 2 n - 1 = 9$
$\Rightarrow \quad n = 5 \mathrm{s}$
So, total time taken by the ball to hit the ground is 5 s. So,
$\begin{array}{r l} & h = \frac {1}{2} g t ^ {2} \\ \Rightarrow & h = \frac {1}{2} (1 0) (5) ^ {2} = 5 (2 5) = 1 2 5 \mathrm{m} \\ \Rightarrow & h = 1 2 5 \mathrm{m} \end{array}$
Hence, the correct answer is (C).

Let $OP = k$
$O R = k r ^ {2} \mathrm{andsoon}$
$\mathrm{then} v _ {P} ^ {2} = 2 a (O P)\tag{... (1}$
$v _ {Q} ^ {2} = 2 a (O Q)\tag{... (2}$
$v _ {R} ^ {2} = 2 a (O R)\tag{... (3}$
and so on.
Hence, the correct answer is (B).

$\begin{array}{r l} & 0 ^ {2} - u ^ {2} = 2 (- g) H \\ & \Rightarrow \quad u = \sqrt {2 g H} \\ & \Rightarrow \quad h = u t - \frac {1}{2} g t ^ {2} \\ & \Rightarrow \quad h = (\sqrt {2 g H}) t - \frac {1}{2} g t ^ {2} \end{array}\tag{... (1}$
Quadratic in t solve t getting two values for time whose ratio is $\frac{t_{1}}{t_{2}}=\frac{1}{3}$ (given). Substituting values we get
4 h = 3 H
Hence, the correct answer is (A).

$\begin{array} v _ {b} = \text { velocity of boat } = v (\text { say }) \\ & v _ {r} = \text { velocity of river } = \eta v \end{array}$
To start with, we must observe that velocity of river flow is $\eta$ times greater than the velocity of boat. So, the boat has to drift and in this problem we are to minimise the drift. Let, the velocity of the boat make an angle $\theta$ with the river velocity. Then $v_{x}=v\cos\theta$ is the component of the velocity of boat along the river flow, whereas $v_{y}=v\sin\theta$ is the component of the velocity of boat perpendicular to river flow. Due to this the net velocity of boat downstream will become $v_{r} + v_{x} = \eta v + v \cos \theta = V$ (say)
If t is the time taken by the boat to cross the river of width L, then
$t = \frac {L}{v _ {y}} = \frac {L}{v \sin \theta}$
During this time the drift equals (say x)
$\begin{array}{l} x = V t \\ x = \big (\eta v + v \cos \theta \big) \frac {L}{v \sin \theta} \\ x = \big (\eta \mathrm{cosec} \theta + \cot \theta \big) L \end{array}$

To MINIMISE x, we must have
$\begin{array}{l l} & \frac {d x}{d \theta} = 0 \\ & \eta (- \operatorname{cosec} \theta \cot \theta) - \operatorname{cosec} ^ {2} \theta = 0 \\ \Rightarrow & \eta \operatorname{cosec} \theta \cot \theta = - \operatorname{cosec} ^ {2} \theta \\ \Rightarrow & \eta \cot \theta = - \operatorname{cosec} \theta \\ \Rightarrow & \eta \frac {\cos \theta}{\sin \theta} = - \frac {1}{\sin \theta} \\ \Rightarrow & \cos \theta = - \frac {1}{\eta} \\ \Rightarrow & - \cos \theta = \frac {1}{\eta} \\ \Rightarrow & \sin \left(\theta - \frac {\pi}{2}\right) = \frac {1}{\eta} \\ \Rightarrow & \theta - \frac {\pi}{2} = \sin^ {- 1} \left(\frac {1}{\eta}\right) \\ \Rightarrow & \theta = \frac {\pi}{2} + \sin^ {- 1} \left(\frac {1}{\eta}\right) \end{array}$
Hence, the correct answer is (C).

Relative to P, the path of S should be along the line SP. So perpendicular to this path, relative velocity should be zero.
Hence, the correct answer is (C).

$\frac {d v}{d t} = - \alpha v$
... (1)
$\Rightarrow v \frac {d v}{d x} = - \alpha v$
$\Rightarrow \frac {d v}{d x} = - \alpha$
$\Rightarrow \int_ {v _ {0}} ^ {0} d v = - \alpha \int_ {0} ^ {s} d x$
$\Rightarrow 0 - v _ {0} = - \alpha s$
$\Rightarrow \quad s = \frac {v _ {0}}{\alpha}$
Also from (1),
${\frac {d v}{v}} = - \alpha d t$
$\Rightarrow \int_ {v _ {0}} ^ {v} \frac {d v}{v} = - \alpha \int_ {0} ^ {t} d t$
$\Rightarrow v = v _ {0} e ^ {- \alpha t}$
$v = 0, \mathrm{when} t \to \infty$
... (2)
$\mathrm{Also}, v = \frac {v _ {0}}{2}$
$\Rightarrow \frac {v _ {0}}{2} = v _ {0} e ^ {- \alpha t}$
$\Rightarrow e ^ {\alpha t} = 2$
$\Rightarrow t = \frac {\log_ {e} (2)}{\alpha} = \frac {7}{1 0 \alpha}$
Hence, (A), (C) and (D) are correct.
$\left\{\because \log_ {e} (2) = 0. 7 \right\}$

$s = ut + \frac{1}{2} at^{2}$
$\Rightarrow v _ {a v} = \frac {2 u + a t}{2} = \frac {u + (u + a t)}{2}$
$\Rightarrow v _ {a v} = \frac {u + v}{2}$
Hence, (A) and (D) are correct.