Kinematics-1D

Here, $\vec{v}=K\left(y\hat{i}+x\hat{j}\right)$
$\Rightarrow \vec {v} = K y \hat {i} + K x \hat {j}\tag{... (1}$
Since, $\vec{v} = \frac{dx}{dt}\hat{i} +\frac{dy}{dt}\hat{j}$
... (2)
Equating equations (1) and (2), we get
${\frac {d x}{d t}} = K y {\mathrm{and}} {\frac {d y}{d t}} = K x$
$\Rightarrow \frac {d y}{d x} = \frac {d y}{d t} \times \frac {d t}{d x}$
$\Rightarrow \frac {d y}{d x} = \frac {K x}{K y} = \frac {x}{y}\tag{... (3}$
Integrating both sides of the above equation, we get
$\int y d y = \int x d x$
$\Rightarrow \quad \frac {y ^ {2}}{2} = \frac {x ^ {2}}{2} + \text { constant }$
$\Rightarrow \quad y ^ {2} = x ^ {2} + \text { constant }$
Hence, the correct answer is (A).

$\vec{v} = \vec{u} +\vec{a} t$
$\Rightarrow \vec {v} = (3 \hat {i} + 4 \hat {j}) + (0. 4 \hat {i} + 0. 3 \hat {j}) \times 1 0$
$\Rightarrow \vec {v} = (3 + 4) \hat {i} + (4 + 3) \hat {j}$
$\Rightarrow \quad | \vec {v} | = \sqrt {4 9 + 4 9} = \sqrt {9 8} = 7 \sqrt {2} \mathrm{units}$
Hence, the correct answer is (B).

Statement-II, is the mathematical definition of relative velocity. However, it does not explain Statement-I correctly. The correct explanation of Statement-I is due to visual perception of motion. The object appears to be moving faster, when its angular velocity is greater w.r.t. observer.
Hence, the correct answer is (B).

Given line have positive intercept but negative slope. So its equation can be written as
$v = - m x + v _ {0}\tag{...(1}$
where $m = \tan \theta = \frac{v_0}{x_0}$
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 equation (1) we get
$\frac {d v}{d t} = - m \big (- m x + v _ {0} \big) = m ^ {2} x - m v _ {0}$
$\Rightarrow a = m ^ {2} x - m v _ {0}$
i.e. the graph between a and x should have positive slope but negative intercept on a-axis. So graph (A) is correct.
Hence, the correct answer is (A).

Distance travelled in $t^{th}$ second is,
$s _ {t} = u + \frac {a}{2} (2 t - 1)$
Since, u = 0, so
$\Rightarrow \quad \frac {s _ {n}}{s _ {n + 1}} = \frac {a n - \frac {1}{2} a}{a (n + 1) - \frac {1}{2} a} = \frac {2 n - 1}{2 n + 1}$
Hence, the correct answer is (C).

The area under acceleration time graph gives change in velocity. As acceleration is zero at the end of 11 s

$\Rightarrow v_{\mathrm{max}} = \mathrm{Area~of~}\Delta OAB$
$\Rightarrow v _ {\max} = \frac {1}{2} \times 1 1 \times 1 0 = 5 5 m s ^ {- 1}$
Hence, the correct answer is (B).

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

|Average Velocity| = $\left|\frac{Displacement}{time}\right| = \frac{AB}{time} = \frac{2}{1} = 2 \, ms^{-1}$
Hence, the correct answer is (B).

$x = a \cos(pt)$
$y = b \sin (p t)$
$\Rightarrow \frac {x ^ {2}}{a ^ {2}} + \frac {y ^ {2}}{b ^ {2}} = 1$
$\Rightarrow$ Trajectory is an Ellipse
$\Rightarrow \quad \ddot {x} = - a p ^ {2} \cos (\omega t)$
$\Rightarrow \quad \ddot {x} = - p ^ {2} x$
Similarly $\ddot{y} = -p^{2}y$
So, acceleration is directed towards the focus.
Hence, (A), (B) and (C) are correct.

For Q
Since acceleration due to gravity has no component along AB. So motion of the particle along AB is non-accelerated motion with uniform velocity v. If $t_{Q}$ is the time taken by particle Q to go from A to B, then
$t _ {Q} = \frac {A B}{v}\tag{... (1}$
For P
Since motion of particle P from A to C is accelerated and that from C to B is retarded. So from A to C the horizontal component of velocity gradually increases to attain a maximum value at the lowest point which further decreases gradually to attain the same value v at point B.
That is the motion from A to B to C experiences a horizontal velocity which has a value more than v. So $t_{Q} > t_{P}$
Hence, the correct answer is (A).

The body starts from rest at x=0 and then again comes to rest at x=1. It means initially acceleration is positive and then negative.
So, we can conclude that $\alpha$ cannot remains positive for all t in the interval $0 \leq t \leq 1$ i.e., $\alpha$ must change sign during the motion.
Hence, (A) and (D) are correct.

$v_{b}$ = velocity of boat in still water = 5 kmh $^{-1}$
$v_{br} =$ velocity of boat w.r.t. river
$v_{r} =$ velocity of river
$v _ {b r} = \frac {1 \mathrm{km}}{\left(\frac {1 5}{6 0}\right) h} = 4 \mathrm{kmh} ^ {- 1}$
By Pythagoras Theorem
$v _ {r} ^ {2} + v _ {b r} ^ {2} = v _ {b} ^ {2}$

$\Rightarrow v _ {r} ^ {2} = v _ {b} ^ {2} - v _ {b r} ^ {2}$
$\Rightarrow v _ {r} ^ {2} = 5 ^ {2} - 4 ^ {2}$
$\Rightarrow v _ {r} = 3 \mathrm{kmh} ^ {- 1}$
Hence, the correct answer is (B).

To swim across the river to a point directly opposite in the shortest time, the man should swim at an angle $\theta = \sin^{-1}\left(\frac{5}{10}\right) = 30^{\circ}$ towards the West of North.
Hence, the correct answer is (C).

In the right angle $\Delta PQR$ , $l^{2} = c^{2} + y^{2}$

c = constant, / and y are variable
Differentiating this equation with respect to time, we get
$2 l \frac {d l}{d t} = 0 + 2 y \frac {d y}{d t}$
Here, $-\frac{dy}{dt} = v_M$
$\Rightarrow \quad \frac {l}{y} = \frac {1}{\cos \theta}$
and $\frac{-dl}{dt} = U$
Hence, $v_{M} = \frac{U}{\cos\theta}$
Hence, the correct answer is (C).

$\vec{a}_{av}=\frac{\Delta\vec{v}}{\Delta t}=\frac{\vec{v}_{f}-\vec{v}_{i}}{\Delta t}$
$\Delta \vec{v} = 5\sqrt{2}\mathrm{ms}^{-1}$ in North-West direction.

$\vec{a}_{av}=\frac{5\sqrt{2}}{10}=\frac{1}{\sqrt{2}}\ ms^{-2}$ (in North-West direction)
Hence, the correct answer is (C).

MISCONCEPTION
Taking east direction as positive, we have
$u = + 9 \mathrm{ms} ^ {- 1} \text { and } a = - 3 \mathrm{ms} ^ {- 2}$
$s _ {5 ^ {t h}} = u + \frac {\alpha}{2} (2 (5) - 1)$
$\Rightarrow s _ {5 ^ {t h}} = 9 - \frac {2}{2} (9) = 0$
Actually this is not the correct answer, because we must observe that the particle reverses its direction of motion (v=0) at t=4.5 s which lies in the specified interval. So, we will proceed other way round.
Since the total distance is
d s = v d t
$\Rightarrow \quad s _ {\mathrm{total}} = \int v d t$
$\Rightarrow \quad s _ {\mathrm{total}} = \left| \int_ {4} ^ {4 \cdot 5} v d t \right| + \left| \int_ {4 \cdot 5} ^ {5} v d t \right|$
$\mathrm{Now} \int v d t = \int (u + a t) d t = u t + \frac {1}{2} a t ^ {2}$
$\begin{array}{r l r} & {\Rightarrow} & {s _ {\mathrm{total}} = \bigg | 9 (4. 5 - 4) + \frac {1}{2} (- 2) \bigg | \bigg [ \bigg (\frac {9}{2} \bigg) ^ {2} - (4) ^ {2} \bigg ] +} \\ & & {\qquad \bigg | 9 (5 - 4. 5) + \frac {1}{2} (- 2) \bigg [ (5) ^ {2} - \bigg (\frac {9}{2} \bigg) ^ {2} \bigg ] \bigg |} \\ & {\Rightarrow} & {s _ {\mathrm{total}} = | 4. 5 - 4. 2 5 | + | 4. 5 - 4. 2 5 | = 0. 5 \mathrm{m}} \end{array}$
Hence, the correct answer is (B).

$\begin{array}{r l} & {a = \frac {d v}{d t} = 8 - 2 t} \\ & {\Rightarrow a = 8 - 2 (5)} \\ & {\Rightarrow a = - 2 \mathrm{ms} ^ {- 2}} \end{array}$
Hence, the correct answer is (D).

For constant speed, $a_{T} = 0$
$\begin{array}{r l} \Rightarrow & a = a _ {C} = a _ {N} = \frac {V _ {T} ^ {2}}{R} \\ \Rightarrow & 0. 8 g = \frac {v ^ {2}}{(2 0 0)} \\ \Rightarrow & v ^ {2} = (0. 8) (1 0) (2 0 0) \\ \Rightarrow & v = 4 0 \mathrm{ms} ^ {- 1} \\ \Rightarrow & v = 4 0 \times \frac {1 8}{5} \mathrm{kmh} ^ {- 1} \\ \Rightarrow & v = 1 4 4 \mathrm{kmh} ^ {- 1} \end{array}$
Hence, the correct answer is (C).

Let the balls A and B collide at the point P a distance h from the ground. Let u be the velocity of launch of B and the launch velocity of A is zero. At the point P, final velocity of A is
$v _ {A} = \sqrt {2 g (H - h)}\tag{... (1}$

and that of $B$ is
$v _ {B} = \sqrt {u ^ {2} - 2 g h}\tag{... (2}$
Also, for the balls to collide at P, time taken by A to reach P equals time taken by B to reach P. So we have
$(H - h) = \frac {1}{2} g t ^ {2}\tag{... (3}$
$h = u t - \frac {1}{2} g t ^ {2}\tag{... (4}$
$(3), t = \sqrt {\frac {2 (H - h)}{g}}$
From
Substitute in (4), we get
$\begin{array}{r l} & h = u \sqrt {\frac {2 (H - h)}{g}} - \frac {1}{2} g \frac {2 (H - h)}{g} \\ \Rightarrow & h = u \sqrt {\frac {2 (H - h)}{g}} - (H - h) \\ \Rightarrow & h = u \sqrt {\frac {2 (H - h)}{g}} - H + h \\ \Rightarrow & u = H \sqrt {\frac {g}{2 (H - h)}} \end{array}\tag{... (5}$
Substitute (5) in (2), we get
$v _ {B} = \sqrt {\frac {g H ^ {2}}{2 (H - h)} - 2 g h}\tag{... (6}$
$\begin{array}{r l} & {\mathrm{Since} v _ {A} = 2 v _ {B}} \\ & {\Rightarrow v _ {A} ^ {2} = 4 v _ {B} ^ {2}} \\ & {\Rightarrow 2 g (H - h) = \frac {4 H ^ {2} g}{2 (H - h)} - 8 g h} \\ & {\Rightarrow 2 g H - 2 g h = \frac {2 H ^ {2} g}{H - h} - 8 g h} \\ & {\Rightarrow H - h = \frac {H ^ {2}}{H - h} - 4 h} \\ & {\Rightarrow \frac {H ^ {2}}{H - h} - H = 3 h} \\ & {\Rightarrow \frac {H ^ {2} - H ^ {2} + H h}{H - h} = 3 h} \\ & {\Rightarrow \frac {H}{H - h} = 3} \\ & {\Rightarrow H = 3 H - 3 h} \\ & {\Rightarrow 2 H = 3 h} \\ & {\Rightarrow h = \frac {2 H}{3}} \end{array}$
Hence, the correct answer is (D).

$\tan\beta=\frac{v_{wind}\sin\theta}{v_{boat}+v_{wind}\cos\theta}$

$\Rightarrow \quad \tan \beta = \frac {7 2 \sin (1 3 5 ^ {\circ})}{5 1 + 7 2 \cos (1 3 5 ^ {\circ})}$
$\Rightarrow \tan \beta \rightarrow \infty$
$\Rightarrow \quad \beta \approx 9 0 ^ {\circ}$
i.e., direction of flag is towards east
Hence, the correct answer is (A).

$L = \frac{1}{2} gt^2 -\frac{1}{2} g(t - T)^2$
Solve $t=\frac{T}{2}+\frac{L}{gT}$
Hence, the correct answer is (B).

Let the point $B$ be situated at a distance $x$ below $A$ . If the particle reaches $B$ in time $t$ , then
$x = \frac {1}{2} g t ^ {2}\tag{... (1}$

For AC, $x + h = \frac{1}{2} g(t + 2)^{2}$
$\ldots(2)$
For AD, $x + 2h = \frac{1}{2} g(t + 3)^{2}$
... (3)
Solving (1), (2) and (3), we get $t = 0.5\mathrm{s}$
Hence, the correct answer is (B).

At the maximum height, v = 0
$\Rightarrow 0 ^ {2} - u ^ {2} = 2 (- g) (h)$
$\begin{array}{r l} \Rightarrow & u = \sqrt {2 g h} \\ \Rightarrow & u = \sqrt {2 (1 0) (2 0)} = 2 0 \mathrm{ms} ^ {- 1} \end{array}$
If t be the time taken by the ball to reach the maximum height, then
0 = u - g t
$\Rightarrow 0 = 2 0 - (1 0) t$
$\Rightarrow t = 2 \mathrm{s}$
So, time taken by the ball to return to juggler's hand is 4 s. To maintain a proper distance between them, the balls must be thrown up at an interval of
$\Delta t = \frac {4}{4} s = 1 \mathrm{s}$
Hence, the correct answer is (B).

C

Let hail fall along the vertical with a velocity $\vec{v}$ . In the reference frame fixed to the car, the angle of incidence of hailstones on the windscreen is equal to the angle of reflection. The velocity of the hailstone before striking the windscreen is $\vec{v}-\vec{v}_{1}$ (by triangle law of vectors). Further the hailstones bounce vertically upwards (from the viewpoint of the driver) after reflection from wind screen of the car. If angle of incidence and hence the angle of reflection both equal $\beta_{1}$ then $\alpha+2\beta_{1}=90^{\circ}$ .

Further $\tan\alpha=\frac{v}{v_{1}}$
$\Rightarrow \quad \tan (9 0 - 2 \beta_ {1}) = \frac {v}{v _ {1}}$
$\Rightarrow \quad \frac {v}{v _ {1}} = \cot (2 \beta_ {1}) \quad \{\text { for 1st car } \}$
Similarly for second car
$\Rightarrow \quad \frac {v}{v _ {2}} = \cot (2 \beta_ {2})$
$\frac {v _ {1}}{v _ {2}} = \frac {\cot (2 \beta_ {2})}{\cot (2 \beta_ {1})} = \frac {\cot (3 0)}{\cot (6 0)} = \frac {\sqrt {3}}{1 / \sqrt {3}} = 3$
Hence, the correct answer is (A).