Kinematics-1D

Ans. (4)
$\vec {r} = 1 5 t ^ {2} \hat {\imath} + (4 - 2 0 t ^ {2}) \hat {j}$
$\vec {v} = \frac {d \vec {r}}{d t} = 3 0 t \hat {\imath} + (- 4 0 t) \hat {j}$
$\vec {a} = \frac {d \vec {v}}{d t} = 3 0 \hat {\imath} - 4 0 \hat {j}$
$| \vec {a} | = 5 0 m / s ^ {2}$

The arrangement of the ships is shown in the Figure (not drawn to scale). Given that

$\vec {v} _ {A} = 3 0 \hat {i} + 5 0 \hat {j} \mathrm{kmhr} ^ {- 1}, \vec {r} _ {A} = 0 \hat {i} + 0 \hat {j}$
$\vec {v} _ {B} = \left(- 1 0 \hat {i}\right) \mathrm{kmhr} ^ {- 1}, \vec {r} _ {B} = \left(8 0 \hat {i} + 1 5 0 \hat {j}\right) \mathrm{km}$
$\Rightarrow \vec {r} _ {B A} = (8 0 \hat {i} + 1 5 0 \hat {j}) \mathrm{km}$
$\Rightarrow \vec {v} _ {B A} = \vec {v} _ {B} - \vec {v} _ {A} = - 1 0 \hat {i} - 3 0 \hat {i} - 5 0 \hat {i} = - 4 0 \hat {i} - 5 0 \hat {j}$
The time after which distance between the ships is minimum is given by
$t = \frac {\vec {r} _ {B A} \cdot \vec {v} _ {B A}}{\left| \vec {v} _ {B A} \right| ^ {2}}$
$\Rightarrow t = \frac {\left(8 0 \hat {i} + 1 5 0 \hat {j}\right) \cdot \left(- 4 0 \hat {i} - 5 0 \hat {j}\right)}{\left| - 4 0 \hat {i} - 5 0 \hat {j} \right| ^ {2}}$
$\Rightarrow t = \left(\frac {3 2 0 0 + 7 5 0 0}{4 1 0 0}\right) \mathrm{hr}$
$\Rightarrow t = \frac {1 0 7 0 0}{4 1 0 0} \mathrm{hr} = 2. 6 \mathrm{hr}$
Hence, the correct answer is (D).

Since $a =$ constant, $u = 0$
$\Rightarrow v = a t$
Since particle starts from the origin
$\Rightarrow x = \frac {1}{2} a t ^ {2}$
So, (I), (II) and (IV) are correct graphs. Hence, the correct answer is (C).

Drawing the relative velocity diagram, we get

$\sin \theta = \frac {v _ {r}}{v _ {s r}} = \frac {1}{2}$
$\Rightarrow \quad \theta = 3 0 ^ {\circ}$
$\Rightarrow \quad \phi = 9 0 + \theta = 1 2 0 ^ {\circ}$
Hence, the correct answer is (D).

$\vec{r}=15t^{2}\hat{i}+4\hat{j}-20t^{2}\hat{j}$
$\Rightarrow \frac {d \vec {r}}{d t} = 3 0 t \hat {i} - 4 0 t \hat {j}$
$\Rightarrow \frac {d ^ {2} \vec {r}}{d t ^ {2}} = 3 0 \hat {i} - 4 0 \hat {j}$
$\Rightarrow \left| \frac {d ^ {2} \vec {r}}{d t ^ {2}} \right| = 5 0 \mathrm{ms} ^ {- 2}$
Hence, the correct answer is (A).

$x = at + bt^{2} - ct^{3}$
Velocity $\dot{x} = \frac{dx}{dt}$ is given by
$\dot {x} = a + 2 b t - 3 c t ^ {2}$
Acceleration $\ddot{x} = \frac{d^2x}{dt^2}$ is given by
$\ddot {x} = 2 b - 6 c t$
For
$\ddot {x} = 0, t = + \frac {b}{3 c}$
$\Rightarrow \quad v = \dot {x} = a + 2 b \left(\frac {+ b}{3 c}\right) - 3 c \left(\frac {b ^ {2}}{3 c \times 3 c}\right)$
$\Rightarrow v = - \frac {b ^ {2}}{3 c} + \frac {2 b ^ {2}}{3 c} + a = a + \frac {b ^ {2}}{3 c}$
Hence, the correct answer is (B).

$v^{2} = u^{2} - 2as$
$\begin{array}{l l} \Rightarrow & v ^ {2} = (1) ^ {2} - (2) \left(\frac {2 . 5 \times 1 0 ^ {- 2}}{2 0 \times 1 0 ^ {- 3}}\right) \left(\frac {2 0}{1 0 0}\right) \\ \Rightarrow & v ^ {2} = 1 - \frac {1}{2} \\ \Rightarrow & v = \frac {1}{\sqrt {2}} \mathrm{ms} ^ {- 1} = 0. 7 \mathrm{ms} ^ {- 1} \end{array}$
Hence, the correct answer is (B).

$v = \frac {d x}{d t} = b \sqrt {x}\tag{... (1}$
$\Rightarrow \int_ {0} ^ {x} \frac {d x}{\sqrt {x}} = \int_ {0} ^ {\tau} b d t$
$\Rightarrow 2 \sqrt {x} = b \tau\tag{... (2}$
$\Rightarrow v = b \left(\frac {b \tau}{2}\right) = \frac {b ^ {2} \tau}{2}$
Hence, the correct answer is (C).

Since $\frac{dx}{dt} = ky$ and $\frac{dy}{dt} = kx$
$\Rightarrow \frac {d y}{d x} = \frac {x}{y}$
$\Rightarrow y d y = x d x$
$\Rightarrow \int y d y = \int x d x$
$\Rightarrow \frac {y ^ {2}}{2} = \frac {x ^ {2}}{2} + \mathrm{constant}$
$\Rightarrow y ^ {2} - x ^ {2} = \mathrm{constant}$
$\Rightarrow y ^ {2} = x ^ {2} + \mathrm{constant}$
Hence, the correct answer is (C).

$\begin{array}{r l} & t _ {A} = t _ {B} - t \\ & \Rightarrow v _ {A} = a _ {1} (t _ {B} - t) = a _ {2} t _ {B} + v \\ & \Rightarrow S = \frac {1}{2} a _ {1} (t _ {B} - t) ^ {2} = \frac {1}{2} a _ {2} t _ {B} ^ {2} \\ & \Rightarrow t _ {B} \left(1 - \sqrt {\frac {a _ {2}}{a _ {1}}}\right) = t \end{array}\tag{... (1}$
... (2)
Solving equations (1) and (2), we get
$v = t \sqrt {a _ {1} a _ {2}}$
Hence, the correct answer is (D).

$v _ {x} = \frac {d x}{d t} = - a \omega \sin (\omega t)$
$v _ {y} = \frac {d y}{d t} = a \omega \cos (\omega t)$
$v _ {z} = \frac {d z}{d t} = a \omega$
$\Rightarrow \quad v = \sqrt {v _ {x} ^ {2} + v _ {y} ^ {2} + v _ {z} ^ {2}} = \sqrt {2} a \omega$
Hence, the correct answer is (B).

Since displacement is the area under v-t graph. So
$\begin{array}{r l} & \text { Area } = \frac {1}{2} \times 2 \times 2 + 2 \times 2 + 3 \times 1 \\ & \Rightarrow \text { Displacement } = 9 \mathrm{m} \\ & \text { Hence, the correct answer is (A). } \end{array}$

When moving in same direction
$t _ {1} = \frac {\ell_ {1} + \ell_ {2}}{v _ {1} - v _ {2}}$
When moving in opposite direction
$t _ {2} = \frac {\ell_ {1} + \ell_ {2}}{v _ {1} + v _ {2}}$
$\Rightarrow \frac {t _ {1}}{t _ {2}} = \frac {v _ {1} + v _ {2}}{v _ {1} - v _ {2}} = \frac {8 0 + 3 0}{8 0 - 3 0} = \frac {1 1}{5}$
Hence, the correct answer is (B).

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

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

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

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

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

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

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

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

If we take the position of ship ‘A’ as origin then positions and velocities of both ships can be given as
$\begin{array}{r l} & {\vec {v} _ {A} = (3 0 \hat {i} + 5 0 \hat {j}) k m / h r} \\ & {\vec {v} _ {B} = - 1 0 \hat {i} k m / h r} \\ & {\vec {r} _ {A} = 0 \hat {i} + 0 \hat {j}} \\ & {\vec {r} _ {B} = (8 0 \hat {i} + 1 5 0 \hat {j}) k m} \end{array}$
Time after which
distance between them
will be minimum

$t = \frac {\vec {r} _ {B A} . \vec {v} _ {B A}}{| \vec {v} _ {B A} | ^ {2}}$
where $\vec{r}_{BA} = (80\hat{i} + 150\hat{j})km$
$\hat {v} _ {B A} = - 1 0 \hat {i} - (3 0 \hat {i} + 5 0 \hat {j})$
$\begin{array}{l} v _ {B A} = - 1 0 i - (3 0 i + 5 0 j) \\ (- 4 0 \hat {i} - 5 0 \hat {j}) k m / h r \\ \therefore t = - \frac {(8 0 \hat {i} + 1 5 0 \hat {j}) . (- 4 0 \hat {i} - 5 0 \hat {j})}{| - 4 0 \hat {i} - 5 0 \hat {j} | ^ {2}} \\ = \frac {3 2 0 0 + 7 5 0 0}{4 1 0 0} h r = \frac {1 0 7 0 0}{4 1 0 0} h r = 2. 6 h r s. \end{array}$

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

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



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