Kinematics-1D

Ans. (ABCD)

$y = \frac{x^2}{2}$ At $t = 0, \left. \begin{array}{l} x = 0, y = 0 \\ u = 1 \end{array} \right\}$ given $y = \frac{x^2}{2}$ $\frac{dy}{dt} = \frac{1}{2}.2x\frac{dx}{dt} \Rightarrow v_y = xv_x$ Differentiate w.r.t. time $a_y = \frac{dx}{dt}.V_x + xa_x$ $a_y = v_x^2 + xa_x$ Option:
(A) If $a_x = 1$ and particle is at origin
(x = 0, y = 0) $a_y = v_x^2$ $a_y = 1^2 = 1$ At origin, at t = 0 sec
Speed = 1 (given)
(B) Option $a_y = v_x^2 + xa_x$ Given in option B, $a_x = 0$ $\Rightarrow a_y = v_x^2$ If $a_x = 0, v_x = \text{constant} = 1$ , (all the time) $\Rightarrow a_y = 1^2 = 1$ (all the time)
(C) At $t = 0, x = 0, v_y = xv_x$ Speed = 1 $v_y = 0$ $v_x = 1$ (D) $a_y = v_x^2 + xa_x$ $v_y = xv_x$ $a_x = 0$ (given in D option) $\Rightarrow a_y = v_x^2$ If $a_x = 0 \Rightarrow V_x = \text{constant initially } (v_x = 1)$ $\Rightarrow a_y = 1^2 = 1$ At $t = 1 \sec$ $v_y = 0 + a_y × t = 1 × 1 = 1$ $\tan \theta = \frac{v_y}{v_x} = x$ (θ → angle with x axis) $\tan \theta = \frac{v_y}{v_x} = \frac{1}{1} = 1$ θ = 45°

$x = a \cos(pt)$
$y = b \sin (p t)$
$\Rightarrow \frac {x ^ {2}}{a ^ {2}} + \frac {y ^ {2}}{b ^ {2}} = 1$
$\Rightarrow$ 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.

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.

A,D

A,B,D

A,B,C,D

B,C,D

B,C,D

A,B,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.

Acceleration $= \frac{dv}{dt} = \dot{v} = 0 + k\dot{x}$
$\Rightarrow \quad \dot {v} = a = k v = k (v _ {0} + k x)$
$\left\{\because \dot {x} = \frac {d x}{d t} = v \right\}$
Further,
$a = \frac {d v}{d t} = k v$
$\Rightarrow \frac {d v}{d t} = k v$
$\Rightarrow \frac {d v}{v} = k d t$
$\Rightarrow \int_ {v _ {0}} ^ {v _ {1}} \frac {d v}{v} = k \int_ {0} ^ {t} d t$
$\Rightarrow t = \frac {1}{k} \mathrm{log} _ {e} \left(\frac {v _ {1}}{v _ {0}}\right)$
Since, $v = v_{0} + kx$ . Hence slope of velocity displace-
ment curve is $\frac{dv}{dx}=k$
Hence, (A), (B) and (C) are correct.

The speed with which the lower end of the rod moves is
$v _ {x} = \frac {d x}{d t}$
$\Rightarrow \quad v _ {x} = \frac {d y}{d t} \left(\frac {d x}{d y}\right)$
$\Rightarrow \frac {d x}{d y} = \frac {- y}{\sqrt {l ^ {2} - y ^ {2}}}$
$\mathrm{Since}, x = \sqrt {l ^ {2} - y ^ {2}}$
$\Rightarrow v _ {x} = \frac {- y}{\sqrt {l ^ {2} - y ^ {2}}} \frac {d y}{d t}$
$\Rightarrow v _ {x} = \frac {y}{\sqrt {l ^ {2} - y ^ {2}}} \left(- \frac {d y}{d t}\right)$
$\Rightarrow v _ {x} = \frac {y v}{\sqrt {l ^ {2} - y ^ {2}}}$
$\mathrm{At} y = 0, v _ {x} = 0$
Hence, (C) and (D) are correct.

Since $v^{2}=v_{x}^{2}+v_{y}^{2}$
$\Rightarrow 2 v \frac {d v}{d t} = 2 v _ {x} \left(\frac {2 v _ {x}}{d t}\right) + 2 v _ {y} \left(\frac {d v _ {y}}{d t}\right)$
$\Rightarrow \frac {d v}{d t} = \frac {v _ {x} a _ {x} + v _ {y} a _ {y}}{v} = \frac {\vec {a} \cdot \vec {v}}{v}$
This is also equal to the projection of $\vec{a}$ along $\vec{v}$ . Hence, (C) and (D) are correct.

$x_{t}=t^{3}-9t^{2}+6t$
$v = \frac {d x _ {t}}{d t} = (3 t ^ {2} - 1 8 t + 6) \mathrm{cms} ^ {- 1}$
For body to be at rest v = 0
$\Rightarrow t = (3 \pm \sqrt {7}) s$
$\Rightarrow t _ {1} = (3 - \sqrt {7}) s \mathrm{and} t _ {2} = (3 + \sqrt {7}) s$
Displacement of particle in travelling from 1st zero of velocity to second zero is -74 cm and v < 0 for
$\left(3 - \sqrt {7}\right) < t < \left(3 + \sqrt {7}\right)$
Hence, (A), (C) and (D) are correct.

The $v - t$ graph for the two cases are shown here For Case A, we have
$v _ {\mathrm{max}} = a _ {1} t _ {1} = a _ {2} t _ {2}\tag{... (1}$
For Case B, we have
$v _ {\mathrm{max}} ^ {\prime} = (2 a _ {1}) t _ {1} = a _ {2} t _ {3}\tag{... (2}$
From these equations, we get
$v _ {\mathrm{max}} ^ {\prime} = 2 v _ {\mathrm{max}} \mathrm{and} t _ {3} = 2 t _ {2}$


$\begin{array}{l} \text {Also,} s _ {1} = (\text {Area under the curve OAB}) \\ \Rightarrow \quad s _ {1} = \frac {1}{2} v _ {\max} (t _ {1} + t _ {2}) \\ \text {and} s _ {2} = (\text {Area under the curve OCD}) \\ \Rightarrow \quad s _ {2} = \frac {1}{2} v _ {\max} ^ {\prime} (t _ {1} + t _ {3}) \\ \Rightarrow \quad s _ {2} = \frac {1}{2} (2 v _ {\max}) (t _ {1} + 2 t _ {2}) \end{array}$
So, the average velocity in Case A, is
$v _ {1} = \frac {s _ {1}}{t _ {1} + t _ {2}} = \frac {v _ {\mathrm{max}}}{2}$
The average velocity in Case B, is
$v _ {2} = \frac {s _ {2}}{t _ {1} + t _ {3}} = \frac {1}{2} v _ {\mathrm{max}} ^ {\prime} = v _ {\mathrm{max}} \qquad \left\{\because v _ {\mathrm{max}} ^ {\prime} = 2 v _ {\mathrm{max}} \right\}$
So, we observe that $v_{2}=2v_{1}$ and $2s_{1} < s_{2} < 4s_{1}$ Hence, (B) and (C) are correct.

$\begin{array}{r l} & {\Rightarrow \frac {d v}{d t} = - \alpha v} \\ & {\Rightarrow v \frac {d v}{d x} = - \alpha v} \\ & {\Rightarrow \int_ {v _ {0}} ^ {0} d v = - \alpha \int_ {0} ^ {x _ {0}} d x} \\ & {\Rightarrow - v _ {0} = - \alpha x _ {0}} \\ & {\Rightarrow x _ {0} = \frac {v _ {0}}{\alpha}} \\ & {\mathrm{Further,} \frac {d v}{d t} = - \alpha v} \\ & {\Rightarrow \int_ {v _ {0}} ^ {v} \frac {d v}{v} = - \alpha \int_ {0} ^ {t} d t} \\ & {\Rightarrow \log_ {e} v | _ {v _ {0}} ^ {v} = - \alpha t | _ {0} ^ {t}} \\ & {\Rightarrow v = v _ {0} e ^ {- \alpha t}} \end{array}$
For $t\to \infty$ , $v\to 0$
Hence particle continues to move for a long time span. Hence, (A) and (B) are correct.

For AC, $v^{2}-(7)^{2}=2al$
... (1)
$\text{For } CB, \quad (17)^2 - v^2 = 2al \tag{2}$
$\text{From (1) and (2), we get:}$
$v^2 = \frac{(7)^2 + (17)^2}{2}$
$\Rightarrow v^2 = 169$
$\Rightarrow v = 13\text{ ms}^{-1}$
$\text{Let } t_1 \text{ be the time taken to go from } A \text{ to } C \text{, and } t_2 \text{ be the time taken to go from } C \text{ to } B.$
$l = \left(\frac{13 + 7}{2}\right) t_1 \quad (\text{for } AC) \tag{3}$
$l = \left(\frac{13 + 17}{2}\right) t_2 \quad (\text{for } CB) \tag{4}$
$\Rightarrow 10t_1 = 15t_2 \implies \frac{t_1}{t_2} = \frac{3}{2}$

When wind blows along the line AB,
$t = t _ {A \to B} + t _ {B \to A}$
$\Rightarrow t = \frac {l}{v + v _ {0}} + \frac {l}{v _ {0} - v}$
$\Rightarrow t = \frac {2 l v _ {0}}{v _ {0} ^ {2} - v ^ {2}}$

If wind blows perpendicular to AB
$t = t _ {A \to B} + t _ {B \to A}$
In both the cases, we have
$v ^ {\prime} = \sqrt {v _ {0} ^ {2} - v ^ {2}}$
$t _ {A \rightarrow B} = \frac {l}{\sqrt {v _ {0} ^ {2} - v ^ {2}}} \mathrm{and}$
$t _ {B \rightarrow A} = \frac {l}{\sqrt {v _ {0} ^ {2} - v ^ {2}}}$
$\Rightarrow t = \frac {2 l}{\sqrt {v _ {0} ^ {2} - v ^ {2}}}$
If the wind were not present then total time taken for the trip would have been $t = \frac{2l}{v_{0}}$
i.e. the total time for the trip increases because of the presence of wind.
Hence, (A), (B) and (D) are correct.

$a = 4 - 2x$
$\Rightarrow v \frac {d v}{d x} = 4 - 2 x$
$\Rightarrow v d v = (4 - 2 x) d x$
$\Rightarrow \int_ {0} ^ {v} v d v = \int_ {0} ^ {x} (4 - 2 x) d x$
$\Rightarrow \frac {v ^ {2}}{2} = 4 x - x ^ {2}$
$\Rightarrow v ^ {2} = 8 x - 2 x ^ {2}\tag{... (1}$
From (1), we observe that v=0, when x=0 and $x=4$ m.
$\mathrm{Now}, a = 0$
$\Rightarrow 4 - 2 x = 0$
$\Rightarrow x = 2 \mathrm{m}$
So, this happens to be the mean position for the particle to oscillate.
Hence, (B) and (C) are correct.

Distance travelled by motor bike at t = 18 s
$s _ {\mathrm{bike}} = s _ {1} = \frac {1}{2} (1 8) (6 0) = 5 4 0 \mathrm{m}$
Distance travelled by car at t = 18 s
$s _ {\mathrm{car}} = s _ {2} = (1 8) (4 0) = 7 2 0 \mathrm{m}$
Therefore, separation between them at t=18 s is 180 m.
Let separation between them decreases to zero at time t beyond 18 s.
Hence, $s_{\mathrm{bike}} = 540 + 60t$ and $s_{\mathrm{car}} = 720 + 40t$
$s _ {\mathrm{car}} - s _ {\mathrm{bike}} = 0$
$\Rightarrow 7 2 0 + 4 0 t = 5 4 0 + 6 0 t$
$\Rightarrow \quad t = 9 \mathrm{s} \text { beyond } 1 8 \mathrm{s}$
$\Rightarrow t = (18 + 9)s = 27s$ from start
and distance travelled by both is $s_{bike} = s_{car} = 1080 \, m$ . Hence, (A) and (C) are correct.

$\left(a_{S}\right)_{HOR} = a_{0} - a\cos \theta$
$\left(a _ {S}\right) _ {V E R} = a _ {0} \sin \theta$

Hence, (B) and (C) are correct.

Since particle is decelerated, Hence
$a = - \alpha \sqrt {v}$
$\Rightarrow \frac {d v}{d t} = - \alpha \sqrt {v}\tag{... (1}$
$\Rightarrow \frac {d v}{d x} \times \frac {d x}{d t} = - \alpha \sqrt {v}$
$\Rightarrow \quad \frac {v}{\sqrt {v}} d v = - \alpha d x$
$\Rightarrow \sqrt {v} d v = - \alpha d x$
Integrating both sides, we have
$\int_ {v _ {0}} ^ {0} v ^ {1 / 2} d v = - \alpha \int_ {0} ^ {x _ {0}} d x$
$x _ {0} = \frac {2}{3} \frac {v _ {0} ^ {3 / 2}}{\alpha}$
Further, from (1), we have
$\Rightarrow v ^ {- 1 / 2} d v = - \alpha d t$
Integrating,
$\int_ {v _ {0}} ^ {0} v ^ {- 1 / 2} d v = - \alpha \int_ {0} ^ {t _ {0}} d t$
$\Rightarrow t _ {0} = 2 \frac {\sqrt {v _ {0}}}{\alpha}$
Hence, (A) and (D) are correct.

$v_{av}=\frac{s_{1}+s_{2}+s_{3}}{t_{1}+t_{2}+t_{3}}$
$\begin{array}{r l} \Rightarrow & 2 0 = \frac {s _ {1} + s _ {2} + s _ {3}}{2 5} \\ \Rightarrow & s _ {1} + s _ {2} + s _ {3} = 5 0 0 \end{array}\tag{... (1}$
Since car accelerated from $0 \rightarrow v_{max}$ in time $t_{1}$ with acceleration of $5 ms^{-2}$ and it decelerated from $v_{max} \rightarrow 0$ in time $t_{3}$ with a retardation of $5 ms^{-2}$ hence $t_{1} = t_{3}$ and $s_{1} = s_{3}$ . Hence,
$t_2 = 25 - 2t_1$
$\Rightarrow s_1 = s_3 = \frac{1}{2}(5)t_1^2 \quad \text{and} \quad v_{\text{max}} = 5t_1$
$\Rightarrow s_2 = v_{\text{max}}t_2 = 5t_1t_2$
$\tag{2}$

Substituting values in (1) and using (2) we get, a quadratic in $t_{1}$ which on solving gives $t_{1}=5$ s.
Hence, $t_2 = 15\mathrm{s}$ and $v_{\mathrm{max}} = 25\mathrm{ms}^{-1}$
Hence, (A), (B), (C) and (D) are correct.

B,D