Kinematics-1D

$v_{x} = \frac{dx}{dt} = -a\omega \sin (\omega t)$
$v _ {y} = \frac {d y}{d t} = a \omega \cos (\omega t)$
$a _ {x} = \frac {d v _ {x}}{d t} = - a \omega^ {2} \cos (\omega t)$
$a _ {y} = \frac {d v _ {y}}{d t} = - a \omega^ {2} \sin (\omega t)$
So, acceleration is $\sqrt{a_x^2 + a_y^2}$
... (1)
$\Rightarrow \mathrm{Acceleration} = a \omega^ {2}\tag{... (1}$
Hence, the correct answer is (C).

For $0 \leq t \leq 2s$
$a _ {1} = 2 \mathrm{ms} ^ {- 2}$
$\Rightarrow \frac {d v _ {1}}{d t} = 2$
$\Rightarrow \int_ {0} ^ {v _ {1}} d v _ {1} = 2 \int_ {0} ^ {t} d t$
$\Rightarrow v _ {1} = 2 t$
$\Rightarrow \frac {d x _ {1}}{d t} = 2 t$
$\Rightarrow d x _ {1} = 2 t d t$
... (2)
$\Rightarrow \int_ {0} ^ {x _ {1}} d x _ {1} = 2 \int_ {0} ^ {2} t d t$
$\begin{array}{r l} \Rightarrow & x _ {1} = 2 \frac {t ^ {2}}{2} \bigg | _ {0} ^ {2} \\ \Rightarrow & x _ {1} = 4 \mathrm{m} \end{array}$
$\begin{array}{r l} & a _ {2} = 1 \mathrm{ms} ^ {- 2} \\ \Rightarrow & \frac {d v _ {2}}{d t} = 1 \end{array}$
$\{\because \mathrm{of}(1)\}$
$\Rightarrow \int_ {4} ^ {v _ {2}} d v ^ {2} = \int_ {2} ^ {t} d t\tag{... (2}$
$\Rightarrow v _ {2} - 4 = t - 2$
$\Rightarrow v _ {2} = 2 + t$
$\Rightarrow \frac {d x _ {2}}{d t} = 2 + t$
$\Rightarrow \int_ {4} ^ {x _ {2}} d x _ {2} = \int_ {2} ^ {4} (2 + t) d t$
$\Rightarrow x _ {2} - 4 = 2 (2) + \frac {1}{2} \left(4 ^ {2} - 2 ^ {2}\right)$
$\Rightarrow x _ {2} - 4 = 4 + 6$
$\Rightarrow x _ {2} = 1 4 \mathrm{m}$
Now $v\Bigg|_{t = 4} = 6\mathrm{ms}^{-1}$
With this initial velocity and a retardation of $1.5 \, ms^{-2}$ , the particle will come to rest in a time t (say). Then
$\begin{array}{r l} & 0 = 6 - (1. 5) t \\ \Rightarrow & t = 4 \mathrm{s} \\ \text {So, for a further 4 s,} \frac {d v _ {3}}{d t} = - 1. 5 \\ & d v _ {3} = - 1. 5 d t \\ \Rightarrow & \int_ {6} ^ {v _ {3}} d v _ {3} = - 1. 5 \int_ {4} ^ {t} d t \\ \Rightarrow & v _ {3} - 6 = - 1. 5 (t - 4) \\ \Rightarrow & v _ {3} - 6 = - 1. 5 t + 6 \\ \Rightarrow & v _ {3} = - 1. 5 t + 1 2 \\ \Rightarrow & \frac {d x _ {3}}{d t} = - 1. 5 t + 1 2 \\ \Rightarrow & \int_ {1 4} ^ {x} d x _ {3} = - 1. 5 \int_ {4} ^ {8} t d t + 1 2 \int_ {4} ^ {8} d t \\ \Rightarrow & x - 1 4 = - 1. 5 \left(\frac {t ^ {2}}{2} \Bigg | _ {4} ^ {8}\right) + 1 2 \left(t \Bigg | _ {4} ^ {8}\right) \\ \Rightarrow & x - 1 4 = - 1. 5 (3 2 - 8) + 1 2 (4) \\ \Rightarrow & x - 1 4 = 1 2 \\ \Rightarrow & x = 2 6 \mathrm{m} \end{array}$
Please be careful while taking the limits in each interval.
Hence, the correct answer is (B).

$80 \, kmh^{-1} NE \equiv 80 \left( \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \right) = \vec{v}_{A}$
$6 0 \mathrm{kmh} ^ {- 1} S E \equiv 6 0 \left(\frac {1}{\sqrt {2}} \hat {i} - \frac {1}{\sqrt {2}} \hat {j}\right) = \vec {v} _ {B}$
Since $\vec{v}_{A / B} =$ velocity of $A$ relative to $B = \vec{v}_A - \vec{v}_B$
$\Rightarrow \vec {v} _ {A / B} = \vec {v} _ {A} - \vec {v} _ {B} = 1 0 \sqrt {2} \hat {i} + 7 0 \sqrt {2} \hat {j}$
$\Rightarrow \quad \theta = \tan^ {- 1} \left(\frac {1}{7}\right).$
Hence, the correct answer is (D).

$\begin{array}{l} s = \frac {t ^ {2}}{2} + \frac {t ^ {3}}{3} \\ v _ {T} = \frac {d s}{d t} = t + t ^ {2} \\ \Rightarrow a _ {T} = 1 + 2 t \\ \Rightarrow a _ {T} | _ {t = 2 s} = 5 \mathrm{ms} ^ {- 2} \\ \text { Now } a _ {C} = \frac {v _ {T} ^ {2}}{r} = \frac {\left(v _ {T} | _ {t = 2}\right) ^ {2}}{r} \\ \Rightarrow a _ {C} = \frac {(6) ^ {2}}{3} = 1 2 \mathrm{ms} ^ {- 2} \\ \text { Since } a ^ {2} = a _ {C} ^ {2} + a _ {T} ^ {2} \\ \Rightarrow a ^ {2} = 1 4 4 + 2 5 \\ \Rightarrow a = 1 3 \mathrm{ms} ^ {- 2} \\ \text { Hence, the correct answer is (D) } \end{array} \tag {34.}$

$\begin{array}{v = \sqrt {2 g h}} \\ & {\Rightarrow v = \sqrt {2 (1 0) 5}} \\ & {\Rightarrow v = 1 0 \mathrm{ms} ^ {- 1}} \\ & {t = \sqrt {\frac {2 h}{g}}} \\ & {\Rightarrow t = \sqrt {\frac {2 (5)}{1 0}} = 1 \mathrm{s}} \end{array}$
So, total time to hit the bottom from the surface is $t' = 5 - 1 = 4$ s
$\Rightarrow d = v t ^ {\prime}$
$\Rightarrow d = (1 0) (4)$
$\Rightarrow d = 4 0 \mathrm{m}$
Hence, the correct answer is (D).

Whenever a particle separated from a body moving with some velocity and having some acceleration, then the particle separated always retains the velocity of the body but does not retain its acceleration. So, in this case we have the velocity of the balloon at t = 8 s is
$v = 0 + (1. 2 5) (8) = 1 0 \mathrm{ms} ^ {- 1} \mathrm{and}$
$s _ {\mathrm{balloon}} = \frac {1}{2} (1. 2 5) (6 4) = 4 0 \mathrm{m}$
When released the stone has a velocity of $10 \, ms^{-1}$ up. So, for the stone, we have
$s = u t + \frac {1}{2} a t ^ {2}$
$\begin{array}{r l} \Rightarrow & - 4 0 = (1 0) t + \frac {1}{2} (- 1 0) t ^ {2} \\ \Rightarrow & - 4 = t - \frac {t ^ {2}}{2} \\ \Rightarrow & - 8 = 2 t - t ^ {2} \\ \Rightarrow & t ^ {2} - 2 t - 8 = 0 \\ \Rightarrow & t ^ {2} - 4 t + 2 t - 8 = 0 \\ \Rightarrow & t (t - 4) + 2 (t - 4) = 0 \\ \Rightarrow & t = - 2 \textrm {s o r} t = 4 \textrm {s} \end{array}$
Hence, the correct answer is (C).

$\begin{array}\quad v = \sqrt {2 g h} \text { and } v ^ {\prime} = \sqrt {2 g (h + x)} \\ \Rightarrow \quad v ^ {\prime} = \sqrt {2 g h} \left(1 + \frac {x}{h}\right) ^ {\frac {1}{2}} \\ \text { Since } x \ll h \\ \therefore \quad v ^ {\prime} = v \left(1 + \frac {x}{2 h}\right) \\ \therefore \quad \text { Increase in velocity } = \frac {v x}{2 h} \\ \text { Hence, the correct answer is (A). } \end{array}$

$ 2=(v\cos\theta-10)t $ and
$3=(v\sin\theta)t$


$(v\cos \theta - 10)$
$\Rightarrow t = \frac {1}{1 0} \left(\frac {3}{\tan \theta} - 2\right)$
$\Rightarrow \quad t = \frac {1}{1 0} \left(\frac {3}{\left(\frac {3}{4}\right)} - 2\right)$
$\Rightarrow t = \frac {1}{1 0} (4 - 2)$
Hence, the correct answer is (A).

Since slope of v-t graph gives acceleration, so we observe that at all the instants acceleration of P is greater than Q.
Hence, the correct answer is (D).

$\begin{array} Since, t = \sqrt {\frac {2 h}{g}} \text { and } t ^ {\prime} = \sqrt {\frac {2 (h + x)}{g}} \\ \Rightarrow \quad t ^ {\prime} = \sqrt {\frac {2 h}{g}} \left(1 + \frac {x}{h}\right) ^ {\frac {1}{2}} \\ \Rightarrow \quad t ^ {\prime} - t \approx \frac {x t}{2 h} \\ \text { Hence, the correct answer is (C). } \end{array}$

$a = -0.6 - 0.001v^{2}$
$\begin{array}{l} v \frac {d v}{d x} = - 0. 6 - 0. 0 0 1 v ^ {2} \\ \Rightarrow \int_ {6 0} ^ {0} \frac {v d v}{0 . 6 + 0 . 0 0 1 v ^ {2}} = - \int_ {0} ^ {s} d x \\ \Rightarrow \frac {1 0 0 0}{2} \int_ {6 0} ^ {0} \frac {2 v d v}{v ^ {2} + 6 0 0} = - s \\ \Rightarrow s = - \frac {1 0 0 0}{2} \log_ {e} (v ^ {2} + 6 0 0) \Big | _ {6 0} ^ {0} \\ \Rightarrow s = - 5 0 0 \log_ {e} \left[ \frac {6 0 0}{(6 0) ^ {2} + 6 0 0} \right] \\ \Rightarrow s = - 5 0 0 \log_ {e} \left(\frac {6 0 0}{4 2 0 0}\right) \\ \Rightarrow s = 5 0 0 \log_ {e} (7) \\ \Rightarrow s = 5 0 0 (1. 9 4 6) \\ \Rightarrow s = 9 7 3 \mathrm{m} \\ \text {Hence, the correct answer is (C).} \end{array}$

$s _ {1} = 4 = \frac {1}{2} \big (v _ {\mathrm{max}} \big) t _ {1}$
$s _ {2} = 9 6 = v _ {\mathrm{max}} t _ {2}$

Since $t_{1} + t_{2} = 10.4$ s
$\Rightarrow \frac {8}{v _ {\mathrm{max}}} + \frac {9 6}{v _ {\mathrm{max}}} = 1 0. 4$
$\Rightarrow \frac {8 + 9 6}{v _ {\mathrm{max}}} = 1 0. 4$
$\Rightarrow v _ {\mathrm{max}} = \frac {1 0 4}{1 0 . 4} = 1 0 \mathrm{ms} ^ {- 1}$
Hence, the correct answer is (A).

$\vec {v} = \dot {\vec {r}} = \frac {d \vec {r}}{d t} = - a \omega \sin (\omega t) \hat {i} + a \omega \cos (\omega t) \hat {j} + b \hat {k}$
$\left| \vec {v} \right| = \sqrt {a ^ {2} \omega^ {2} + b ^ {2}}$
Distance moved by the particle in one full turn of helix is given by
$s = \left| \vec {v} \right| T = \frac {2 \pi}{\omega} \sqrt {a ^ {2} \omega^ {2} + b ^ {2}}$
Hence, the correct answer is (B).

Area under a v-t graph gives the distance travelled. So
$\begin{array}{r l} & s = (2 0) (6) + \frac {1}{2} (2 0 + 4 5) 3 + \frac {1}{2} (3) (4 5) \\ \Rightarrow & s = 1 2 0 + 1 6 5 \\ \Rightarrow & s = 2 8 5 \mathrm{m} \end{array}$
Hence, the correct answer is (C).

$\vec {v} _ {w} = \frac {v}{\sqrt {2}} \hat {i} + \frac {v}{\sqrt {2}} \hat {j}$
$\begin{array}{l} \vec {v} _ {m} = (a t) \hat {j} \\ \Rightarrow \quad \vec {v} _ {w m} = \vec {v} _ {w} - \vec {v} _ {m} = \frac {v}{\sqrt {2}} \hat {i} + \left(\frac {v}{\sqrt {2}} - a t\right) \hat {j} \end{array}$

It appears due east when, the $y$ component vanishes. So
$\begin{array}{r l} & {\frac {v}{\sqrt {2}} - a t = 0} \\ {\Rightarrow} & {t = \frac {v}{\sqrt {2} a}} \end{array}$
Hence, the correct answer is (C).

From graph
$v _ {f} ^ {2} = \left(\text { final velocity }\right) ^ {2} = 9 0 0 \mathrm{km} ^ {2} \mathrm{h} ^ {- 2}$
$\begin{array}{l} v _ {i} ^ {2} = \left(\text {initial velocity}\right) ^ {2} = 3 6 0 0 \mathrm{km} ^ {2} \mathrm{h} ^ {- 2} \\ s = \text {Distance covered} = 0. 6 \mathrm{km} \end{array}$
Since, $v_{f}^{2} - v_{i}^{2} = 2as$ we get $a = -2250\mathrm{kmh}^{-2}$
Hence, the correct answer is (C).

$\begin{array}{r l} & {a = \frac {v - u}{t} = \frac {1 0 - 2 5}{3} = - 5 \mathrm{ms} ^ {- 2}} \\ & {s = u t + \frac {1}{2} a t ^ {2}} \\ & {\Rightarrow s = 2 5 (3) + \frac {1}{2} (- 5) (9)} \\ & {\Rightarrow s = 7 5 - 2 2. 5} \\ & {\Rightarrow s = 5 2. 5 \mathrm{m}} \end{array}$
Acceleration is constant, so (C) happens to be incorrect. Hence, the correct answer is (A).

$\begin{array}{l} v _ {a v} = \frac {l + l}{\frac {l}{2 v _ {1}} + \frac {l}{2 v _ {2}}} \\ \Rightarrow v _ {a v} = \frac {2 v _ {1} v _ {2}}{v _ {1} + v _ {2}} = 4 8 \mathrm{kmh} ^ {- 1} \end{array}$
Hence, the correct answer is (D).

$\tan \beta = \frac {v _ {g i r l} + v _ {r a i n} \sin \alpha}{v _ {r a i n} \cos \alpha}$
$\Rightarrow \tan \beta = \frac {8 + 1 0 \sin \alpha}{1 0 \cos \alpha}$
Hence, the correct answer is (B).

The particle strikes the point B, when the velocity of the particle with respect to the platform is along AB i.e., the component of relative velocity along AD must be zero.
$\begin{array}{l l} \Rightarrow & v - 2 v \cos \theta = 0 \\ \Rightarrow & \cos \theta = \frac {1}{2} \\ \Rightarrow & \theta = 6 0 ^ {\circ} \end{array}$
Hence, the correct answer is (C).

In this problem, we must check that the distance travelled by the elevator while accelerating plus the distance travelled by the elevator while decelerating must equal 100 m. If this is not satisfied then the remaining distance must have been travelled with a uniform speed equal to the maximum speed attained by the elevator at the end of accelerated motion.
Hence, the correct answer is (B).

Let v be the velocity of the river and u be the velocity of swimmer with respect to river. Then
$t _ {1} = \frac {2 d}{\sqrt {u ^ {2} - v ^ {2}}}\tag{... (1}$
Further
$t _ {2} = \frac {d}{u + v} + \frac {d}{u - v} = \frac {2 u d}{u ^ {2} - v ^ {2}}\tag{... (2}$
and $t_{3}=\frac{2d}{u}$
... (3)
From (1), (2) and (3), we get
$t _ {1} ^ {2} = t _ {2} t _ {3}$
Hence, the correct answer is (D).

On passing through a plank of thickness x, the velocity of bullet becomes $\frac{19}{20}u$ , where u is the initial velocity.
$\Rightarrow \left(\frac {1 9 u}{2 0}\right) ^ {2} - u ^ {2} = 2 a x\tag{... (1}$
Let the bullet further pass through n planks. Hence total number of planks = $(n+1)$
$0 ^ {2} - u ^ {2} = 2 a (n + 1) x$ .....(2)
Solving we get, $(n+1)\approx11$
Hence, the correct answer is (B).

$v=\frac{6\ km}{2(1\ hr)}=3\ kmh^{-1}$
Hence, the correct answer is (A).