Kinematics-1D

$v = -\left(\frac{v_{0}}{x_{0}}\right)x + v_{0}$
$a = \frac {v d v}{d x}$
$a = \left[ - \left(\frac {v _ {0}}{x _ {0}}\right) x + v _ {0} \right] \left[ - \frac {v _ {0}}{x _ {0}} \right]$
$a = \left(\frac {v _ {0}}{x _ {0}}\right) ^ {2} x - \frac {v _ {0} ^ {2}}{x _ {0}}.$

Option (b) represent correct graph for particle moving with constant acceleration, as for constant acceleration velocity time graph is straight line with positive slope and x-t graph should be an opening upward parabola.

For $0 \leq x \leq 200$
v = m x + C
$v = \frac {1}{5} x + 1 0$
$a = \frac {v d v}{d x} = \left(\frac {x}{5} + 1 0\right) \left(\frac {1}{5}\right)$

$a = \frac{x}{25} + 2 \Rightarrow$ Straight line till $x = 200$
$\mathrm{For} x > 2 0 0$
$v = \mathrm{constant}$
$\Rightarrow a = 0$
Hence most appropriate option will be (a).

$a t + 2 a t ^ {2} = u t + 2 a t ^ {2} + \frac {1}{2} a t ^ {2} \Rightarrow u = \frac {a t}{2}$
For overall motion,
$v = u + 3 a t = \frac {a t}{2} + 3 a t = \frac {7 a t}{2} = 7 u \quad \left(\because u = \frac {a t}{2}\right)$

→ (+)

$V _ {A} = 3 6 k m / h r = 1 0 m / s, V _ {B} = - 7 2 k m / h r = - 2 0 m / s$
$V _ {M A} = - 1. 8 k m / h r = - 0. 5 m / s$
$V _ {m a n, B} = V _ {m a n, A} + V _ {A, B} = V _ {m a n, A} + V _ {A} - V _ {B}$
= - 0. 5 + 1 0 - (- 2 0) = - 0. 5 + 3 0 = 2 9. 5 m / s.

Given $\vec{u} = 5\hat{j} m / s, \vec{a} = 10\hat{i} + 4\hat{j}$ , final coordinate $(20, y_0)$ in time $t$
$S _ {x} = u _ {x} t + \frac {1}{2} a _ {x} t ^ {2}$

$v ^ {2} = u ^ {2} + 2 a s$
$\Rightarrow \sqrt {4 g h} = - \sqrt {2 g h} + g t.$
$v ^ {2} = 2 g h + 2 g h$
$v = \sqrt {4 g h}$
$\Rightarrow t = \sqrt {\frac {4 h}{g}} + \sqrt {\frac {2 h}{g}} \Rightarrow 3. 4 \sqrt {\frac {h}{g}}$

S= area of graph=Total distance covered $=\frac{1}{2}\times4\left(1+\frac{13}{3}\right)+\frac{1}{2}\times2\times\frac{5}{3}=\frac{37}{3}m.$

Rain is falling vertically downwards
$\vec {v} _ {r / m} = \vec {v} _ {r} - \vec {v} _ {m}$
$\tan 6 0 ^ {\circ} = \frac {v _ {r}}{v _ {m}} = \sqrt {3}$

$v _ {r} = v _ {m} \sqrt {3} = v \sqrt {3}$
$v_{r / m}$
Now, $v_{m} = (1 + \beta)v$
and $\theta = 45^{\circ}$
$\tan 4 5 ^ {\circ} = \frac {v _ {r}}{v _ {m}} = 1$

$v _ {r} = v _ {m}$
$v \sqrt {3} = (1 + \beta) v$
$\sqrt {3} = 1 + \beta$
$\Rightarrow \beta = \sqrt {3} - 1 = 0. 7 3.$

$v^{2} = u^{2} + 2gh$

$v^{2} = u^{2} + 2as$

$u = 7 8. 4 \Rightarrow u = 2 6. 1 \mathrm{ms} ^ {- 1}$

When a particle completes one revolution in circular motion, then average displacement travelled by particle is zero.
Hence, average velocity
$= \frac {\text { average displacement }}{\Delta t} = \frac {0}{0} = 0$

Time interval between 8th and 3rd second,
$\Delta t = 8 - 3 = 5 \mathrm{s}, l e. \Delta t = 5 \mathrm{s}$
Change in velocity,
$\Delta v = 2 0 - 0 = 2 0 \mathrm{ms} ^ {- 1}$
$\therefore$ Average acceleration $= \frac{\Delta v}{\Delta t} = \frac{20}{5} = 4\mathrm{ms}^{-2}$

For $A$ and $B$ let time taken by $A$ is $t_0$

Given,
$v _ {A} - v _ {B} = v = (a _ {1} - a _ {2}) t _ {0} - a _ {2} t\tag{....(i}$
$\begin{array}{r l} & x _ {B} = x _ {A} = \frac {1}{2} a _ {1} t _ {0} ^ {2} = \frac {1}{2} a _ {2} (t _ {0} + t) ^ {2} \\ & \Rightarrow \sqrt {a _ {1}}. t _ {0} = \sqrt {a _ {2}} (t _ {0} + t) \\ & \Rightarrow (\sqrt {a _ {1}} - \sqrt {a _ {2}}) t _ {0} = \sqrt {a _ {2}}. t \end{array}\tag{....(ii}$
putting $t_0$ in equation
$\begin{array}{l} v = (a _ {1} - a _ {2}) \frac {\sqrt {a _ {2}} t}{\sqrt {a _ {1}} - \sqrt {a _ {2}}} - a _ {2} t \\ = (\sqrt {a _ {1}} + \sqrt {a _ {2}}) \sqrt {a _ {2}} t - a _ {2} t \\ = \sqrt {a _ {1} a _ {2}} t + a _ {2} t - a _ {2} t \Rightarrow v = \sqrt {a _ {1} a _ {2}} t. \end{array}$

$S = \text{Area under graph} = \left(\frac{1}{2} \times 2 \times 2\right) + (2 \times 2) + (3 \times 1) = 9m.$

$\theta = \omega t = \omega \frac{\pi}{2\omega} = \frac{\pi}{2}$
$\vec {V} _ {A} - \vec {V} _ {B} = \omega R _ {1} (- \hat {i}) - \omega R _ {2} (- \hat {i}) = \omega (R _ {2} - R _ {1}) \hat {i}.$

For swimmer to cross the river straight
$\Rightarrow 4\sin \theta = 2\Rightarrow \sin \theta = \frac{1}{2}\Rightarrow \theta = 30^{0}$
So, angle with direction of river flow = $90^{0} + \theta = 120^{0}$

$x = at + bt^2 - ct^3$
$v = \frac {d x}{d t} = a + 2 b t - 3 c t ^ {2}$
$a = \frac {d v}{d t} = 2 b - 6 c t = 0 \Rightarrow t = \frac {b}{3 c}$
$V _ {\left(a t t = \frac {b}{3 c}\right)} = a + 2 b \left(\frac {b}{3 c}\right) - 3 c \left(\frac {b}{3 c}\right) ^ {2} = a + \frac {b ^ {2}}{3 c}.$

$\vec{r}(t)=15t^{2}\hat{i}+(4-20t^{2})\hat{j}$
$\vec {v} = \frac {d \vec {r}}{d t} = 3 0 t \hat {i} + (- 4 0 t) \hat {j} \Rightarrow \vec {a} = \frac {d \vec {v}}{d t} = 3 0 \hat {i} - 4 0 \hat {j}$
$| \vec {a} | = 5 0 m / s ^ {2}.$

$-(g + \gamma v^2) = \frac{dv}{dt}$
$- d t = \frac {1}{\gamma} \left(\frac {d v}{\frac {g}{\gamma} + v ^ {2}}\right) \Rightarrow - \int_ {0} ^ {t} d t = \frac {1}{\gamma} \int_ {V _ {0}} ^ {0} \frac {d v}{\left(\sqrt {\frac {g}{\gamma}}\right) ^ {2} + v ^ {2}}$
$- t = - \frac {1}{\sqrt {\gamma g}} \tan^ {- 1} \left(\frac {v _ {0}}{\sqrt {\frac {g}{\gamma}}}\right) \Rightarrow t = \frac {1}{\sqrt {\gamma g}} \tan^ {- 1} \left(\sqrt {\frac {\gamma}{g}} v _ {0}\right).$

$v = b\sqrt{x} \Rightarrow \frac{dv}{dt} = \frac{b}{2\sqrt{x}}\frac{dx}{dt}$
$a = \frac {b v}{2 \sqrt {x}} = \frac {b (b \sqrt {x})}{2 \sqrt {x}}$
$\frac {d v}{d t} = a = \frac {b ^ {2}}{2} \Rightarrow v = \frac {b ^ {2}}{2} \tau$

$v = 20m / s$ , $u = 10m / s$
$\sin \theta = \frac {u}{v} = \frac {1 0}{2 0} = \frac {1}{2}$

$\Rightarrow \theta = 3 0 ^ {\circ} \text { west. }$

Given initial velocity $u = 0$ and acceleration is constant $v = 0 + at \Rightarrow v = at$
$x = 0 (t) + \frac {1}{2} a t ^ {2} \Rightarrow x = \frac {1}{2} a t ^ {2}$



Graph (A); (B) and (D) are correct.

According to the question,
For the time duration $ 0 < t < 1 $ s,
the velocity increase from 0 to 6 ms $ ^{-1} $ .
As the direction of field has been reversed, for $ 1 < t < 2 $ s, the velocity firstly decreases from 6 ms $ ^{-1} $ to 0.
Then, for $ 2 < t < 3 $ s, as the field strength is same; the magnitude of acceleration would be same, but velocity increases from 0 to $ -6 \, ms^{-1} $ .

Acceleration of the car,
$| a | = \left| \frac {v - u}{t} \right| = \frac {6 - 0}{1} = 6 \mathrm{ms} ^ {- 2}$
The displacement of the particle is given as
$s = u t + \frac {1}{2} a t ^ {2}$
For $t = 0$ to $t = 1$ s,
$u = 0, a = + 6 \mathrm{ms} ^ {- 2} \Rightarrow s _ {1} = 0 + \frac {1}{2} \times 6 \times (1) ^ {2} = 3 \mathrm{ms}$
For $ t = 1 $s to $ t = 2 $s,
$u = 6 \mathrm{ms} ^ {- 1}, a = - 6 \mathrm{ms} ^ {- 2}$
$\Rightarrow \quad s _ {2} = 6 \times 1 - \frac {1}{2} \times 6 \times (1) ^ {2} = 6 - 3 = 3 \mathrm{m}$
For $t = 2$ to $t = 3\mathrm{s}$,
$u = 0, a = - 6 \mathrm{ms} ^ {- 2} \Rightarrow s _ {3} = 0 - \frac {1}{2} \times 6 \times (1) ^ {2} = - 3 \mathrm{m}$
∴ Net displacement,
$s = s _ {1} + s _ {2} + s _ {3} = 3 \mathrm{m} + 3 \mathrm{m} - 3 \mathrm{m} = 3 \mathrm{m}$
Hence, average velocity
$= \frac {\text { net displacement }}{\text { total time }} = \frac {3}{3} = 1 \mathrm{ms} ^ {- 1}$
Total distance travelled, d = 9 m
Hence, average speed
$= \frac {\text { total distance }}{\text { total time }} = \frac {9}{3} = 3 \mathrm{ms} ^ {- 1}$
Alternate Method
Given condition can be represented through graph also as shown below.

∴ Displacement in three seconds
= Area under the graph
= Area of $\Delta OAY +$ Area of $\Delta AOY =$ Area of $\Delta BCD$
$= \frac {1}{2} \times 1 \times 6 + \frac {1}{2} \times 1 \times 6 - \frac {1}{2} \times 6 \times 1 = 3 \mathrm{m}$
∴ Average velocity = $ \frac{3}{3} = 1ms^{-1} $ .
Total distance travelled, d = 9 m
∴ Average speed = $ \frac{9}{3} $ = 3 ms $ ^{-1} $