Kinematics-1D

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).

In OPTIONS (A), (C) and (D), the given graphs represent uniformly decelerated motion of a particle in a straight line with positive initial velocity. Distance-time graph of such a motion is shown here.

Hence, the correct answer is (B).

Distance travelled by car in 15 s is
$s _ {1} = \frac {1}{2} \times A C \times O C = \frac {1}{2} (4 5) (1 5) = \frac {6 7 5}{2} \mathrm{m}$
Distance travelled by scooter in 15 s is
$s _ {2} = v \times t = 3 0 \times 1 5 = 4 5 0 \mathrm{m}$
Required difference in distance is $(s_{2}-s_{1})=\Delta s$
$\Rightarrow \quad \Delta s = 4 5 0 - \frac {6 7 5}{2} = \frac {2 2 5}{2} = 1 1 2. 5 \mathrm{cm}$
Let the car catch scooter in time t, then
$\begin{array}{r l} & {\frac {6 7 5}{2} + 4 5 (t - 1 5) = 3 0 t} \\ {\Rightarrow} & {3 3 7. 5 + 4 5 t - 6 7 5 = 3 0 t} \\ {\Rightarrow} & {1 5 t = 3 3 7. 5} \\ {\Rightarrow} & {t = 2 2. 5 \mathrm{s}} \end{array}$
Hence, the correct answer is (D).

Using, $v^{2} = u^{2} - 2as$ , we get
$\begin{array}{r l} & 0 = u ^ {2} - 2 a s \\ \Rightarrow & s = \frac {u ^ {2}}{2 a} \\ \Rightarrow & \frac {s _ {1}}{s _ {2}} = \frac {u _ {1} ^ {2}}{u _ {2} ^ {2}} \\ \Rightarrow & s _ {2} = \left(\frac {u _ {2}}{u _ {1}}\right) ^ {2} s _ {1} \\ \Rightarrow & s _ {2} = (2) ^ {2} (4 0) = 1 6 0 \mathrm{m} \\ & \text { Hence, the correct answer is (C). } \end{array}$

Time taken by car to reach at location $P$ from location $Q$ is
$\begin{array}{r l} & t = \frac {Q R}{v} + \frac {R P}{(v / 2)} \\ \Rightarrow & t = \frac {(L - x)}{v} + \frac {2 \sqrt {d ^ {2} + x ^ {2}}}{v} \end{array}$
For t to be minimum, we have

$\begin{array}{l l} & \frac {d t}{d x} = 0 \\ \Rightarrow & \frac {d t}{d x} = \frac {1}{v} (0 - 1) + 2 \times \left(\frac {1}{2}\right) \frac {1}{v} \times \frac {2 x}{\sqrt {d ^ {2} + x ^ {2}}} \\ \Rightarrow & \frac {d t}{d x} = \frac {- 1}{v} + \frac {2 x}{v \sqrt {d ^ {2} + x ^ {2}}} \end{array}$
Since, for minimum value of t
$\begin{array}{r l} & {\frac {d t}{d x} = 0} \\ {\Rightarrow} & {- \frac {1}{v} + \frac {2 x}{v \sqrt {d ^ {2} + x ^ {2}}} = 0} \\ {\Rightarrow} & {1 = \frac {2 x}{\sqrt {d ^ {2} + x ^ {2}}}} \\ {\Rightarrow} & {4 x ^ {2} = d ^ {2} + x ^ {2}} \\ {\Rightarrow} & {3 x ^ {2} = d ^ {2}} \\ {\Rightarrow} & {x = \frac {d}{\sqrt {3}}} \end{array}$
Hence, the correct answer is (B).

Velocity of the body going upwards is given by $v = v_{0} - gt$ , where $v_{0}$ is the initial velocity.
Hence, the graph between velocity and time should be a straight line with negative slope $(g)$ and intercept $v_{0}$ .
Also, during the whole motion, acceleration of the body is constant i.e., slope should be constant.
Hence, the correct answer is (C).

Here, acceleration is given by, a = -c
$\Rightarrow \frac {d v}{d t} = - c$
$\Rightarrow \frac {d x}{d t} \cdot \frac {d v}{d x} = - c$
$\Rightarrow v \frac {d v}{d x} = - c$
$\Rightarrow v d v = - c d x$
$\Rightarrow \frac {v ^ {2}}{2} = - c x + k$
$\Rightarrow x = - \frac {v ^ {2}}{2 c} + \frac {k}{c}$
Hence, the correct answer is (A).

C

Acceleration of car, $a_{C} = 4 \, ms^{-2}$
Acceleration of bus, $a_{B} = 2 \, ms^{-2}$
Initial separation between the bus and car is
$s _ {C B} = 2 0 0 \mathrm{m}$
Acceleration of car with respect to bus is
$a _ {C B} = a _ {C} - a _ {B} = 2 \mathrm{ms} ^ {- 2}$
Initial velocity of car w.r.t. bus is
$u _ {C B} = 0, t = ?$
$\begin{array}{r l} & {\mathrm{Since,} s _ {C B} = u _ {C B} t + \frac {1}{2} a _ {C B} t ^ {2}} \\ & {\Rightarrow 2 0 0 = (0) t + \frac {1}{2} (2) t ^ {2}} \\ & {\Rightarrow t ^ {2} = 2 0 0} \\ & {\Rightarrow t = 1 0 \sqrt {2} \mathrm{s}} \end{array}$
Hence, the correct answer is (C).

Using $h = ut + \frac{1}{2} at^{2}$
For stone 1, we have
$y _ {1} = 1 0 t - \frac {1}{2} g t ^ {2}$
and for stone 2, we have
$y _ {2} = 4 0 t - \frac {1}{2} g t ^ {2}$
Relative position of the second stone with respect to the first,
$\Delta y = y _ {2} - y _ {1} = 4 0 t - \frac {1}{2} g t ^ {2} - 1 0 t + \frac {1}{2} g t ^ {2}$
$\Rightarrow \Delta y = 3 0 t$
After 8 seconds, stone 1 reaches ground, i.e., $y_{1} = -240$ m
$\Rightarrow \quad \Delta y = y _ {2} - y _ {1} = 4 0 t - \frac {1}{2} g t ^ {2} + 2 4 0$
Therefore, it will be a parabolic curve till other stone reaches ground.
Hence, the correct answer is (A).

Time taken by the particle to reach the top most point is,
$t = \frac {u}{g}\tag{... (1}$

Time taken by the particle to reach the ground is
$t ^ {\prime} = n t$
$s = u t + \frac {1}{2} a t ^ {2}$
$\Rightarrow - H = u (n t) - \frac {1}{2} g (n t) ^ {2}$
Using (1), we get
$\begin{array}{r l} & {- H = u \times n \left(\frac {u}{g}\right) - \frac {1}{2} g n ^ {2} \left(\frac {u}{g}\right) ^ {2}} \\ {\Rightarrow} & {- 2 g H = 2 n u ^ {2} - n ^ {2} u ^ {2}} \\ {\Rightarrow} & {2 g H = n u ^ {2} (n - 2)} \end{array}$
Hence, the correct answer is (D).

$\begin{array}{l} \frac {d v}{d t} = - 2. 5 \sqrt {v} \\ \Rightarrow \quad \frac {1}{\sqrt {v}} d v = - 2. 5 d t \end{array}$
On integrating, within limits i.e., $v_{1}=6.25\ ms^{-1}$ to $v_{2}=0$ , we get
$\int_ {6. 2 5} ^ {0} v ^ {- 1 / 2} d v = - 2. 5 \int_ {0} ^ {t} d t$
$\Rightarrow 2 \sqrt {v} \big | _ {6. 2 5} ^ {0} = - (2. 5) t$
$\Rightarrow t = \frac {- 2 \sqrt {6 . 2 5}}{- 2 . 5} = 2 \mathrm{s}$
Hence, the correct answer is (B).