Kinematics-1D

Let v be the velocity of escalator and u be the velocity of the person. Then
$t _ {1} = \frac {l}{u}\tag{... (1}$
Further $t_{2}=\frac{l}{v}$
... (2)
If t be the time taken by him to walk up the escalator, then
$t = \frac {l}{u + v}$
From (1) and (2)
$u = \frac {l}{t _ {1}} \text { and } v = \frac {l}{t _ {2}}$
$\Rightarrow t = \frac {l}{u + v} = \frac {l}{\frac {l}{t _ {1}} + \frac {l}{t _ {2}}}$
$\Rightarrow t = \frac {t _ {1} t _ {2}}{t _ {1} + t _ {2}}$
Hence, the correct answer is (D).

Since both are falling under gravity, so relative acceleration of one stone w.r.t. other is zero. So,
$\Delta x = \left(u _ {1} + u _ {2}\right) t = u t$
$\left\{\because u _ {1} = 0 \right\}$
So, the graph is a straight line passing through the origin.
Hence, the correct answer is (C).

Velocity of wind is $10 \, ms^{-1}$ from South to North i.e.
$\vec {v} _ {\mathrm{wind}} = 1 0 \hat {j}$

But, to the cyclist it appears to blow from the East at $10 \, ms^{-1}$ . So, velocity of wind relative to the cyclist is $10 \, ms^{-1}$ from East to West i.e.
$\vec {v} _ {\mathrm{wind/cyclist}} = - 1 0 \hat {i}$
Since $\vec{v}_{\mathrm{wind / cyclist}} = \vec{v}_{\mathrm{wind}} - \vec{v}_{\mathrm{cyclist}}$
$\Rightarrow - 1 0 \hat {i} = 1 0 \hat {j} - \vec {v} _ {\mathrm{cyclist}}\tag{14.}$
$\Rightarrow \vec {v} _ {\text { cyclist }} = 1 0 \hat {i} + 1 0 \hat {j}$
Hence, the correct answer is (B).

${\frac {v}{t}} = {\frac {v _ {1} - v _ {2}}{t _ {2} - t _ {1}}} = {\frac {\sqrt {2 a _ {1} s} - \sqrt {2 a _ {2} s}}{\sqrt {\frac {2 s}{a _ {2}}} - \sqrt {\frac {2 s}{a _ {1}}}}}$
$\Rightarrow \quad v = \left(\sqrt {a _ {1} a _ {2}}\right) t$
$\Rightarrow v = \sqrt {(2 g) (8 g)} t$
$\Rightarrow \quad v = 4 g t$
Hence, the correct answer is (A).

The graph shows the following v-s relation i.e.,
v = s
$\Rightarrow \frac {d v}{d s} = 1$
$\mathrm{Since} a = \frac {v d v}{d s}$
$\Rightarrow a = v (1)$
$\Rightarrow a = v$
So, a-v graph is again a straight line passing through the origin and inclined to the x-axis at an angle of $45^{\circ}$ . Hence, the correct answer is (C).

According to Law of sines or Lami's Theorem

$\Rightarrow \quad \frac {v _ {A}}{\sin 7 5 ^ {o}} = \frac {v _ {B}}{\sin 6 0 ^ {o}} = \frac {v _ {B / A}}{\sin 4 5 ^ {o}}$
$\Rightarrow v _ {B} = 7 1 7 \mathrm{km} h ^ {- 1}$
Hence, the correct answer is (C).

Since, $h = \frac{1}{2} gt^2$
... (1)
$\mathrm{and} \frac {9 h}{1 6} = \frac {1}{2} g (t - 1) ^ {2}\tag{... (2}$
$\Rightarrow \quad \frac {3}{4} = \frac {t - 1}{t}$
$\Rightarrow 3 t = 4 t - 4$
$\Rightarrow t = 4 \mathrm{s}$
$\Rightarrow h = 8 0 \mathrm{m}$
Hence, the correct answer is (C).

The graph shown is actually for
$s = u t - \frac {1}{2} a t ^ {2}$
Lets see this, by making the LHS a perfect square.
$\begin{array}{r l} \Rightarrow & s = - \frac {a}{2} \bigg (t ^ {2} - \frac {2 u t}{a} \bigg) \\ \Rightarrow & s = - \frac {a}{2} \bigg [ t ^ {2} - \bigg (\frac {2 u}{a} \bigg) t + \frac {u ^ {2}}{a ^ {2}} - \frac {u ^ {2}}{a ^ {2}} \bigg ] \\ \Rightarrow & s = - \frac {a}{2} \bigg [ \bigg (t - \frac {u}{a} \bigg) ^ {2} - \frac {u ^ {2}}{a ^ {2}} \bigg ] \\ \Rightarrow & s - \frac {u ^ {2}}{2 a} = - \frac {a}{2} \bigg (t - \frac {u}{a} \bigg) ^ {2} \\ \Rightarrow & S = - \frac {a}{2} T ^ {2} \end{array}$
where $S=\frac{s-u^{2}}{2a}$ and $T=t-\frac{u}{a}$ and the origin at
$\left(T = 0, S = 0\right) \equiv \left(\frac {u}{a}, \frac {u ^ {2}}{2 a}\right)$
$\begin{array}{r l} & {\mathrm{So,} \frac {u}{a} = 2. 5 \mathrm{and} \frac {u ^ {2}}{2 a} = 1 0 0} \\ & {\Rightarrow \quad \frac {u ^ {2}}{2 a} = \frac {u}{2} \bigg (\frac {u}{a} \bigg) = 1 0 0} \\ & {\Rightarrow \quad \frac {u}{2} (2. 5) = 1 0 0} \\ & {\Rightarrow \quad u = \frac {2 0 0}{2 . 5} = 8 0 \mathrm{ms} ^ {- 1}} \\ & {\mathrm{and} a = \frac {8 0}{2 . 5} = 3 2 \mathrm{ms} ^ {- 2}} \end{array}$
Hence, the correct answer is (C).

$\begin{array} s _ {1} = \frac {1}{2} a t ^ {2} \\ & s _ {2} = (a t) t - \frac {1}{2} a \frac {t ^ {2}}{4} \\ & \Rightarrow \quad s _ {2} = a t ^ {2} - \frac {a t ^ {2}}{8} \\ & \Rightarrow \quad s _ {2} = \frac {7 a t ^ {2}}{8} \end{array}$
Hence, the correct answer is (A).

$v_{i}$ be the initial velocity i.e. velocity at lowest point
$\Rightarrow \quad \overrightarrow {v _ {i}} = u \hat {i}$
$v_{f}$ be the final velocity i.e. velocity when the string becomes vertical. So, we have
(in magnitude)
$\begin{array}{r l} & v _ {f} ^ {2} - v _ {i} ^ {2} = 2 (- g) L \\ \Rightarrow & v _ {f} = \sqrt {u ^ {2} - 2 g L} \quad (\text {in}) \\ & \left[ \begin{array}{c c} \sqrt {u ^ {2} - 2 g L} = v _ {f} \\ \text {O} & \text {L} \\ \text {L} & \text {x} \end{array} \right] \\ \Rightarrow & \vec {v} _ {f} = \left(\sqrt {u ^ {2} - 2 g L}\right) \hat {j} \\ & \Delta \vec {v} = \vec {v} _ {f} - \vec {v} _ {i} = \left(\sqrt {u ^ {2} - 2 g L}\right) \hat {j} - u \hat {i} \\ \Rightarrow & | \Delta \vec {v} | = \sqrt {u ^ {2} - 2 g L + u ^ {2}} = \sqrt {2 (u ^ {2} - g L)} \end{array}$
Hence, the correct answer is (D).

$\begin{array} v = \frac {d x}{d t} = 3 a t ^ {2} + 2 b t + c \\ & \Rightarrow u = v \Big | _ {t = 0} = c \\ & \Rightarrow a = \frac {d v}{d t} = 6 a t + 2 b \\ & \Rightarrow a \Big | _ {t = 0} = 2 b \\ & \Rightarrow \frac {a \Big | _ {t = 0}}{v \Big | _ {t = 0}} = \frac {2 b}{c} \end{array}\tag{... (1}$
Hence, the correct answer is (B).

$\begin{array}a = \frac {P}{m v} \\ & \Rightarrow v \frac {d v}{d x} = \frac {P}{m v} \\ & \Rightarrow v ^ {2} d v = \frac {P}{m} d x \\ & \Rightarrow \int_ {v _ {1}} ^ {v _ {2}} v ^ {2} d v = \frac {P}{m} \int_ {0} ^ {x} d x \end{array}$
$\left\{\because P = \mathrm{constant} \right\}$
$\begin{array}{r l} \Rightarrow & \frac {v ^ {3}}{3} \bigg | _ {v _ {1}} ^ {v _ {2}} = \frac {P}{m} \bigg (x \bigg | _ {0} ^ {x} \bigg) \\ \Rightarrow & v _ {2} ^ {3} - v _ {1} ^ {3} = \frac {3 P x}{m} \\ \Rightarrow & x = \frac {m}{3 P} \big (v _ {2} ^ {3} - v _ {1} ^ {3} \big) \end{array}$
Hence, the correct answer is (C).

$\begin{array}\frac {d v}{d t} = a - b v \\ & \Rightarrow \int_ {0} ^ {v} \frac {d v}{a - b v} = \int_ {0} ^ {t} d t \\ & \Rightarrow - \frac {1}{b} \left[ \log_ {e} (a - b v) - \log_ {e} (a) \right] = t \\ & \Rightarrow \frac {a - b v}{a} = e ^ {- b t} \\ & \Rightarrow v = \frac {a}{b} (1 - e ^ {- b t}) \end{array}$
At t=0 i.e. initially, velocity must be zero (as stated in the Problem) and this condition is met only by Option (A).
Hence, the correct answer is (A).

Initial velocity of dropping = zero
Let $v_{1}$ be velocity at end of 10 s.
$\begin{array}{r l} \Rightarrow & v _ {1} = g t \\ \Rightarrow & v _ {1} = 1 0 0 \mathrm{ms} ^ {- 1} \end{array}$
Distance travelled during this time is
$\begin{array}{r l} & h _ {1} = \frac {v _ {1} ^ {2}}{2 g} = \frac {(1 0 0) ^ {2}}{2 (1 0)} \\ \Rightarrow & h _ {1} = 5 0 0 \mathrm{m} \end{array}$
So, a remaining distance of 2495 - 500 = 1995 m has to be travelled with a retardation of $2.5\mathrm{ms}^{-2}$ . Let the parachutist strike the ground with velocity $v$ . Then
$\begin{array}{r l} & v ^ {2} - v ^ {\prime 2} = 2 a (h - h ^ {\prime}) \\ \Rightarrow & v ^ {2} - (1 0 0) ^ {2} = 2 (- 2. 5) (1 9 9 5) \\ \Rightarrow & v ^ {2} = 1 0 0 0 0 - 9 9 7 5 \\ \Rightarrow & v ^ {2} = 2 5 \end{array}$
$\Rightarrow v = 5 \, \mathrm{ms}^{-1}$ Hence, the correct answer is (A).

$\begin{array} Since s = \left(\frac {u + v}{2}\right) t \\ & \qquad 4 = \left(\frac {1 0 0 0 + 9 0 0 0}{2}\right) t \\ & \Rightarrow t = \frac {8}{1 0 0 0 0} \\ & \Rightarrow t = 8 \times 1 0 ^ {- 4} \mathrm{s} \\ & \text { Hence, the correct answer is (B). } \end{array}$

$\begin{array}{l} s _ {A} = s _ {B} \\ \qquad 8 t + \frac {1}{2} (2) t ^ {2} = 4 t + \frac {1}{2} (4) t ^ {2} \\ \Rightarrow \quad 8 + t = 4 + 2 t \\ \Rightarrow \quad t = 4 \text {s} \\ \text {Since} s _ {A} = 8 t + \frac {1}{2} (2) t ^ {2} \\ \Rightarrow \quad s _ {A} = 8 (4) + (4) ^ {2} \\ \Rightarrow \quad s _ {A} = 4 8 \text {m} \\ \text {Hence, the correct answer is (C).} \end{array}$

$\begin{array}{l} v _ {y} = \frac {d y}{d t} = \frac {1}{2} \frac {d x}{d t} = \frac {1}{2} v _ {x} \\ \Rightarrow \quad v _ {y} = \frac {d y}{d t} = 2 - t \qquad \{\because v _ {x} = 4 - 2 t \} \\ a _ {x} = - 2 \mathrm{ms} ^ {- 2} \text {and} a _ {y} = - 1 \mathrm{ms} ^ {- 2} \\ u _ {x} = v _ {x} \big | _ {t = 0} = 4 \mathrm{ms} ^ {- 1} \text {and} u _ {y} = v _ {y} \big | _ {t = 0} = 2 \mathrm{ms} ^ {- 1} \end{array}$
Since we observe that $a_{x}$ and $a_{y}$ both are negative but $u_{x}$ and $u_{y}$ are positive, so motion is retarded initially and accelerated once the direction of motion is reversed.
Hence, the correct answer is (D).

Let the ball be at height h at time t and time $(t + \Delta t)$ . Then
$h = ut - \frac{1}{2}gt^2$
$h = u(t + \Delta t) - \frac{1}{2}g(t + \Delta t)^2$
$h = \frac{4u^2 - g^2(\Delta t)^2}{8g}$
$u = \frac{1}{2}\sqrt{8gh + g^2(\Delta t)^2}$

$v_{x} = \frac{dx}{dt} = 30\cos (6t)$ and
$v _ {y} = \frac {d y}{d t} = 3 0 \sin (6 t)$
$\Rightarrow v = \sqrt {v _ {x} ^ {2} + v _ {y} ^ {2}} = 3 0 \mathrm{ms} ^ {- 1}$
So, we observe that the speed is constant. Hence distance
x = v t
$\Rightarrow x = (3 0) (4) = 1 2 0 \mathrm{m}$
Hence, the correct answer is (D).

Let the acceleration be A and initial velocity be U. Then
$\begin{array}{r l} & a = U p + \frac {1}{2} A p ^ {2} \\ & a + b = U (p + q) + \frac {1}{2} A (p + q) ^ {2} \\ \Rightarrow & b = U q + \frac {1}{2} A q ^ {2} + A p q \\ & q (1) - p (2) \text {gives} \\ & q a - p b = \frac {1}{2} A p ^ {2} q - \frac {1}{2} A q ^ {2} p - A p ^ {2} q \\ \Rightarrow & q a - p b = - \frac {1}{2} A q ^ {2} p - \frac {1}{2} A p ^ {2} q \\ \Rightarrow & p b - q a = \frac {1}{2} A p q (p + q) \\ \Rightarrow & A = \frac {2 (p b - q a)}{p q (p + q)} = \frac {2 (b p - a q)}{p q (p + q)} \end{array}$
Hence, the correct answer is (D).
Similarly

$v_{x} = \frac{dx}{dt} = -a\omega \sin (\omega t)$
$v _ {y} = \frac {d y}{d t} = a \omega \cos (\omega t)$
$a _ {x} = \frac {d v _ {x}}{d t} = - a \omega^ {2} \cos (\omega t)$
$a _ {y} = \frac {d v _ {y}}{d t} = - a \omega^ {2} \sin (\omega t)$
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).