Kinematics-1D

Let hail fall along the vertical with a velocity $\vec{v}$ . In the reference frame fixed to the car, the angle of incidence of hailstones on the windscreen is equal to the angle of reflection. The velocity of the hailstone before striking the windscreen is $\vec{v}-\vec{v}_{1}$ (by triangle law of vectors). Further the hailstones bounce vertically upwards (from the viewpoint of the driver) after reflection from wind screen of the car. If angle of incidence and hence the angle of reflection both equal $\beta_{1}$ then $\alpha+2\beta_{1}=90^{\circ}$ .

Further $\tan\alpha=\frac{v}{v_{1}}$
$\Rightarrow \quad \tan (9 0 - 2 \beta_ {1}) = \frac {v}{v _ {1}}$
$\Rightarrow \quad \frac {v}{v _ {1}} = \cot (2 \beta_ {1}) \quad \{\text { for 1st car } \}$
Similarly for second car
$\Rightarrow \quad \frac {v}{v _ {2}} = \cot (2 \beta_ {2})$
$\frac {v _ {1}}{v _ {2}} = \frac {\cot (2 \beta_ {2})}{\cot (2 \beta_ {1})} = \frac {\cot (3 0)}{\cot (6 0)} = \frac {\sqrt {3}}{1 / \sqrt {3}} = 3$
Hence, the correct answer is (A).

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).

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