Kinematics-1D

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

$x_{t} = 8t^{2} - 3t^{3}$
$\Rightarrow v _ {t} = 1 6 t - 9 t ^ {2}$
For particle to be at rest $v_{t}=0$
$\Rightarrow 1 6 t - 9 t ^ {2} = 0$
$\Rightarrow t = 0 \mathrm{or} t = \frac {1 6}{9} s$
Substituting $t=\frac{16}{9}$ s in $x_{t}=8t^{2}-3t^{3}$ , we get
$x _ {t} = 8. 4 3 \mathrm{m}$
$\begin{array}{l} \text {Now} v _ {a v} = \frac {x _ {t = 4} - x _ {t = 0}}{4 - 0} \\ \Rightarrow v _ {a v} = \frac {\left[ 8 (4) ^ {2} - 3 (4) ^ {3} \right] - \left[ 8 (0) ^ {2} - 3 (0) ^ {3} \right]}{4} \\ \Rightarrow v _ {a v} = \frac {8 (1 6) - 3 (6 4)}{4} = - 1 6 \mathrm{ms} ^ {- 1} \end{array}$
$\mathrm{Acceleration} = a = \frac {d v _ {t}}{d t} = 1 6 - 1 8 t$
$\Rightarrow a | _ {t = 4 s} = 1 6 - 1 8 (4) = - 5 6 \mathrm{ms} ^ {- 2}$
$\begin{array}{c} (t = 0) A \\ \text { } C \text { } (t = 4 s) \\ \text { } B \text { } (t = 1 6 / 9 s) \\ \text { } 6 4 m \text { } 8. 4 3 m \end{array}$
$\text { Average speed } = \frac {\text { Total Distance }}{\text { Total Time }}$
Total Distance = (64 + 8.43 + 8.43) m = 80.86 m
$\Rightarrow \quad \text { Average speed } = \frac {8 0 . 8 6}{4} = 2 0. 2 1 \mathrm{ms} ^ {- 1}$
Hence, (A), (B), (C) and (D) are correct.

In general, we have $|d\vec{r}| = dr$ and $v \neq \frac{dr}{dt}$ . You can think about these cases by taking an example of circular motion.
Hence, (A), (B) and (D) are correct.

$\begin{array}{r l} A & \xrightarrow {u _ {1} a _ {1}} \xrightarrow {} \xrightarrow {} \xrightarrow {} \xrightarrow {} \xrightarrow {} \xrightarrow {} \xrightarrow {} B \\ & \xleftarrow {} \xleftarrow {} I / 2 \xrightarrow {} \xleftarrow {} \xleftarrow {} \xleftarrow {} I / 2 \xrightarrow {} \xleftarrow {} I \\ & \frac {l}{2} = u _ {1} t + \frac {1}{2} a _ {1} t ^ {2} \\ \text { and } & - \frac {l}{2} = - u _ {2} t + \frac {1}{2} (- a _ {2}) t ^ {2} \\ \Rightarrow & \frac {l}{2} = u _ {2} t + \frac {1}{2} a _ {2} t ^ {2} \end{array}\tag{... (1}$
$\ldots(2)$
Subtracting (1) & (2), we get
$t = 2 \left(\frac {u _ {2} - u _ {1}}{a _ {1} - a _ {2}}\right)\tag{... (3}$
Substituting (3) in (1) or (2) and rearranging, we get
$l = \frac {4 (u _ {2} - u _ {1})}{(a _ {1} - a _ {2}) ^ {2}} (a _ {1} u _ {2} - a _ {2} u _ {1})\tag{... (4}$
Since the particles $P \& Q$ reach the other ends of $A$ and $B$ with equal velocities say $v$
For particle P
$v ^ {2} - u _ {1} ^ {2} = 2 a _ {1} l\tag{... (5}$
For particle Q
$v ^ {2} - u _ {2} ^ {2} = 2 a _ {2} l\tag{... (6}$
Subtracting and then substituting value of $\ell$ and rearranging, we get
$\big (u _ {2} + u _ {1} \big) \big (a _ {1} - a _ {2} \big) = 8 \big (a _ {1} u _ {2} - a _ {2} u _ {1} \big)$
Hence, (A), (B) and (C) are correct.

My dear students, most of you forget to keep in mind the nature of motion, whether accelerated or decelerated, is just not predicted by checking the sign of $\vec{a}$ . It also requires the nature of $\vec{v}$ to be defined or mentioned along with. So keep in mind the following table for describing an accelerated or a retarded motion.

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

Since, $x = 2 + 2t + 4t^{2}$ and $y = 4t + 8t^{2}$
$\begin{array}{r l} \Rightarrow & x = 2 + \frac {y}{2} \\ \Rightarrow & y = 2 x - 4 \end{array}\tag{... (1}$
This is the equation of a straight line
$\frac {d x}{d t} = 2 + 8 t$
So, the motion is uniformly accelerated along a straight line.
Hence, (C) and (D) are correct.

The separation r, between the particle at time t is
$r ^ {2} = (4 - 2 t) ^ {2} + (2 t) ^ {2}\tag{... (1}$
This separation will first decrease and then increase. This r is minimum, when
$\begin{array}{r l} & {\frac {d}{d t} (r ^ {2}) = 0} \\ {\Rightarrow} & {2 (4 - 2 t) (- 2) + 2 (2 t) (2) = 0} \\ {\Rightarrow} & {- 8 + 4 t + 4 t = 0} \\ {\Rightarrow} & {t = 1 \mathrm{s}} \end{array}$

At this time, we have
$\begin{array}{r l} & r _ {\min} ^ {2} = (4 - 2) ^ {2} + (2) ^ {2} \\ \Rightarrow & r _ {\min} ^ {2} = 8 \\ \Rightarrow & r _ {\min} = 2 \sqrt {2} \mathrm{m} \end{array}$
Hence, (B), (C) and (D) are correct.

This problem can be answered if you think about the situation when the particle reverses its direction of motion, because in that case $a \neq 0$ but v = 0, but this is not an always condition.
So, a can be non-zero when v = 0. Another answer is (D), because this situation is just like when the v is constant, then a = 0.
Hence, (B) and (D) are correct.

At time T, the particle reverses its direction of motion, because v=0 and also a=constant. So, (A) and (B) are correct. Also from the graph we observe that the initial and the final speeds of the particle have same value. So, (C) happens to be correct too. Finally, displacement of the particle is actually the area under the v-t graph which is zero. So, (D) is also correct. You can also think about this graph for a body through up with an initial velocity under the influence of gravity. So it will return to the point of start, hence giving zero displacement.
Hence, (A), (B), (C) and (D) are correct.

A,C,D

METHOD I
$v = \sqrt {x}$
$v ^ {2} - 0 ^ {2} = 2 a x$
Please keep this thing in mind that whenever you observe $v^{2} \propto x$ OR $v \propto \sqrt{x}$ OR $x \propto t^{2}$ OR $t \propto \sqrt{x}$ OR $v \propto t$ , then acceleration is constant or the motion is uniformly accelerated.
$\begin{array}{r l} & \text { METHOD II } \\ & v = \sqrt {x} \\ & \Rightarrow \frac {d v}{d t} = \left(\frac {1}{2} x ^ {\frac {1}{2} - 1}\right) \frac {d x}{d t} \\ & \Rightarrow \frac {d v}{d t} = \left(\frac {1}{2 \sqrt {x}}\right) v \\ & \text { But } v = \sqrt {x} \\ & \Rightarrow \frac {d v}{d t} = \frac {1}{2} \\ & \Rightarrow a = \frac {1}{2} \mathrm{ms} ^ {- 2} \end{array}$
Hence, (A), (B) and (C) are correct.

A,B,C