Kinematics-1D

For $AB$
$2 0 0 = \left(\frac {2 0 + v}{2}\right) t _ {1}\tag{... (1}$

For $BC$
$4 0 0 = v t _ {2}$
For CD
$\ldots(2)$
$\begin{array}{r l} 6 0 0 & = \left(\frac {2 0 + v}{2}\right) t _ {3} \\ \Rightarrow & t _ {1} = \frac {1}{3} t _ {3} \\ \text {Since,} & t _ {2} = t _ {1} + t _ {3} = 4 t _ {1} \\ \Rightarrow & 4 0 0 = 4 v t _ {1} \\ \Rightarrow & v t _ {1} = 1 0 0 \end{array}\tag{... (3}$
... (4)
Divide (1) and (4), we have
$\begin{array}{r l} & 2 = \frac {2 0 + v}{2 v} \\ \Rightarrow & 4 v = 2 0 + v \\ \Rightarrow & v = \frac {2 0}{3} \mathrm{ms} ^ {- 1} \\ \Rightarrow & t _ {1} = 1 5 s, t _ {2} = 6 0 s, t _ {3} = 4 5 s \\ \text { Hence }, t = t _ {1} + t _ {2} + t _ {3} \\ \Rightarrow & t = 1 2 0 \mathrm{s} = 2 \text { minute } \\ \text { Hence,the correct answer is(C) }. \end{array}$

From $O \rightarrow A$ we have acceleration varying linearly with time hence
$\begin{array}{r l} & {\Rightarrow a - 0 = \frac {5}{6} (t - 0)} \\ & {\Rightarrow a = \frac {5}{6} t} \\ & {\Rightarrow \frac {d v}{d t} = \frac {5}{6} t} \\ & {\Rightarrow \int_ {0} ^ {v _ {1}} d v = \frac {5}{6} \int_ {0} ^ {t} t d t} \\ & {\Rightarrow v _ {1} = \frac {5}{1 2} t ^ {2}} \\ & {\mathrm{Further} v _ {1} = \frac {d x _ {1}}{d t} = \frac {5}{1 2} t ^ {2}} \\ & {\Rightarrow \int_ {0} ^ {x _ {1}} d x _ {1} = \frac {5}{1 2} \int_ {0} ^ {6} t ^ {2} d t \Rightarrow x _ {1} = \frac {5}{3 6} t ^ {3} \bigg | _ {0} ^ {6}} \\ & {\Rightarrow x _ {1} = \frac {5}{3 6} (6 ^ {3}) = 3 0 \mathrm{m}} \end{array}$
For $A \rightarrow B$ the acceleration is constant and have a value of $5 \, ms^{-2}$ . Now velocity at point A is
$v _ {1} = \frac {5}{1 2} (3 6) = 1 5 \mathrm{ms} ^ {- 1}$
$\Rightarrow x _ {2} = v _ {1} (6) + \frac {1}{2} (5) (6) ^ {2}$
$\Rightarrow x _ {2} = 1 8 0 \mathrm{m}$
So, total distance travelled is $x = 30 + 180 = 210$ m
Hence, the correct answer is (B).

Let u be the initial velocity and a be the acceleration. Then
$\begin{array}{r l} & 2 = u (2) + \frac {1}{2} a (2) ^ {2} \\ \Rightarrow & u + a = 1 \\ & 4. 2 = u (2 + 4) + \frac {1}{2} a (2 + 4) ^ {2} \\ \Rightarrow & 4. 2 = 6 u + 1 8 a \\ \Rightarrow & 0. 7 = u + 3 a \\ \Rightarrow & u + 3 a = 0. 7 \end{array}\tag{... (1}$
... (2)
From (1) and (2), we get
2 a = - 0. 3
$a = - 0. 1 5 \mathrm{ms} ^ {- 2}$
$\mathrm{and} u = 1. 1 5 \mathrm{ms} ^ {- 1}$
$\begin{array}{r l} & v \Big | _ {t = 7 s} = u + 7 a \\ \Rightarrow & v \Big | _ {t = 7 s} = 1. 1 5 - 7 (0. 1 5) \Rightarrow v \Big | _ {t = 7 s} = 1. 1 5 - 1. 0 5 \\ \Rightarrow & v \Big | _ {t = 7 s} = 0. 1 \mathrm{ms} ^ {- 1} \end{array}$
Hence, the correct answer is (A).

For $A \to B$ , $h = \frac{1}{2} gt_1^2$
... (1)
For $A\to C$ , $2h = \frac{1}{2} g(t_1 + t_2)^2$
... (2)
For $A\to D$ , $3h = \frac{1}{2} g(t_1 + t_2 + t_3)^2$
... (3)

Please note that u=0 for all the intervals that start from A.
From (1), (2) and (3), we get
$t _ {1}: t _ {2}: t _ {3} = 1: (\sqrt {2} - 1): (\sqrt {3} - \sqrt {2})$
Hence, the correct answer is (C).

Initially, when both are thrown up simultaneously, then acceleration of A w.r.t. B is zero (as both fall under the influence of gravity). So, relative position versus time plot is a straight line. Now the particle A reaches the ground and stops, so its acceleration becomes zero and hence the relative acceleration of B w.r.t. A is g.
Therefore the relative position vs t graph is now parabolic. So, the variation plot is first linear and then parabolic.
Hence, the correct answer is (D).

Initially let the dog and cat be at A and B respectively with their velocities perpendicular to each other as shown

FIGURE 1: Positions of dog and cat at t = 0
The cat has a velocity constant in magnitude and direction but dog has a velocity continuously aimed at cat i.e. constant in magnitude but varying in direction.
Let at any instant t, v makes an angle $\theta$ with u. Then let us resolve u parallel to v and perpendicular to v as shown in Figure (2). Component parallel to v is $u\cos\theta$ and perpendicular to v is $u\sin\theta$
Relative velocity of dog towards cat is
$(v - u \cos \theta) = - \frac {d x}{d t}$

FIGURE 2: Positions of dog and cat at time t
The negative sign on RHS indicates that the distance between cat and dog decreases if the dog has to catch the cat.
$\begin{array}{l l} \Rightarrow & - d x = (v - u \cos \theta) d t \\ \Rightarrow & - \int_ {\ell} ^ {0} d x = \int_ {0} ^ {t} v d t - \int_ {0} ^ {t} (u \cos \theta) d t \\ \Rightarrow & l = \int_ {0} ^ {t} v d t - \int_ {0} ^ {t} (u \cos \theta) d t \end{array}\tag{... (1}$
For dog to catch the cat we must have
$\binom{\text { Horizontal }}{\text { displacement }} = \binom{\text { Horizontal }}{\text { displacement }}$
$\mathrm{i.e.,} u t = \int_ {0} ^ {t} (v \cos \theta) d t\tag{... (2}$
$\text { From (2) } \int_ {0} ^ {t} (\cos \theta) d t = \frac {u t}{v}$
Substituting in (1), we get
$l = v t - u \left(\frac {u t}{v}\right)$
$\Rightarrow l = \frac {(v ^ {2} - u ^ {2}) t}{v}$
$\Rightarrow t = \frac {v l}{v ^ {2} - u ^ {2}}$
Also for dog to catch the cat v > u.
Hence, the correct answer is (A).

$v^{2} \propto s$ $\Rightarrow$ acceleration is constant.
Hence, the correct answer is (A).

$\begin{array}{l} A B = \left(1 0 0 \times \frac {5}{1 8}\right) t \text {metre} \\ C D = \left(6 0 \times \frac {5}{1 8}\right) t \text {metre} \\ \text {For BC, v_{C} = v_{B} + at} \\ \Rightarrow 6 0 \times \frac {5}{1 8} = 1 0 0 \times \frac {5}{1 8} + a (4) \\ \Rightarrow 4 a = - 4 0 \times \frac {5}{1 8} \\ \Rightarrow a = - \frac {5 0}{1 8} = - \frac {2 5}{9} \mathrm{ms} ^ {- 2} \end{array}$
So the distance BC is
$\begin{array}{l} B C = 1 0 0 \times \frac {5}{1 8} \times 4 + \frac {1}{2} \left(- \frac {2 5}{9}\right) (4) ^ {2} \\ B C = \frac {1 0 0 0}{9} - \frac {2 0 0}{9} = \frac {8 0 0}{9} \mathrm{m} \end{array}$
However $AB + BC + CD = 3000 \mathrm{~m}$
$\begin{array}{r l} \Rightarrow & (1 6 0) \left(\frac {5}{1 8}\right) t + \frac {8 0 0}{9} = 3 0 0 0 \\ \Rightarrow & \frac {4 0 0}{9} t + \frac {8 0 0}{9} = 3 0 0 0 \\ \Rightarrow & 4 0 0 t + 8 0 0 = 2 7 0 0 0 \\ \Rightarrow & 4 0 0 t = 2 6 2 0 0 \\ \Rightarrow & t = 6 5. 5 \mathrm{s} \end{array}$
Hence, the correct answer is (C).

If we take the maximum attainable velocity as $24 \, ms^{-1}$ , then to accelerate with $1 \, ms^{-2}$ (from rest), the time is
$t _ {1} = 2 4 \mathrm{s}$
$\left\{\because 2 4 = 0 + (1) t _ {1} \right\}$

During this time the distance covered is observed to be
$s _ {1} = \frac {1}{2} a _ {1} t _ {1} ^ {2}$
$\Rightarrow s _ {1} = \frac {1}{2} (1) (2 4) ^ {2} = 2 8 8 \mathrm{m} > 2 0 0 \mathrm{m}$
which is not possible.
So, the car must attain a $v_{max} < 24 \, ms^{-1}$
Let that maximum velocity be v. Then
$t _ {1} + t _ {2} = t \mathrm{(say)and} s _ {1} + s _ {2} = 2 0 0 \mathrm{m}$
$\Rightarrow \frac {v}{1} + \frac {v}{4} = t$
$\Rightarrow v = \frac {4 t}{5}\tag{... (1}$
Now $s_1 + s_2 = 200$
$\Rightarrow \frac {v ^ {2}}{2 (1)} + \frac {v ^ {2}}{2 (4)} = 2 0 0$
$\Rightarrow \frac {1 6 t ^ {2}}{5 0} + \frac {1 6 t ^ {2}}{2 0 0} = 2 0 0$
$\Rightarrow 1 6 t ^ {2} \left(\frac {5}{2 0 0}\right) = 2 0 0$
$\Rightarrow t ^ {2} = 5 0 0$
$\Rightarrow t \cong 2 2. 4 \mathrm{s}$
You could have directly applied the formula
$s = \frac {1}{2} \left(\frac {\alpha \beta}{\alpha + \beta}\right) t ^ {2}$
where s = 200 m, t = ?
$\alpha = 1 \mathrm{ms} ^ {- 2} \text { and } \beta = 4 \mathrm{ms} ^ {- 2}$
$\Rightarrow t = 2 2. 4 \mathrm{s}$
Hence, the correct answer is (A).

Let the length of the escalator be $\ell$ . If $u$ be the velocity of person, then
$9 0 = \frac {l}{u}\tag{... (1}$
If v be the velocity of the escalator, then
$6 0 = \frac {l}{v}\tag{... (2}$
If t be the time taken by him to walk up the moving escalator, then
$t = \frac {l}{u + v}\tag{... (3}$
From (1), $u = \frac{l}{90}$
From (2), $v = \frac{l}{60}$
Substituting in (3), we get
$t = \frac {l}{\frac {l}{9 0} + \frac {l}{6 0}}$
$\Rightarrow t = \frac {(6 0) (9 0)}{6 0 + 9 0}$
$\Rightarrow t = 3 6 \mathrm{s}$
Hence, the correct answer is (A).

Let the particles collide at time t.

For particles to collide at C
$\begin{array}{r l} & {\vec {r} _ {B} + \vec {v} _ {B} t = \vec {r} _ {A} + \vec {v} _ {A} t} \\ {\Rightarrow} & {\vec {r} _ {B} - \vec {r} _ {A} = (\vec {v} _ {A} - \vec {v} _ {B}) t} \\ {\Rightarrow} & {| \vec {r} _ {B} - \vec {r} _ {A} | = | \vec {v} _ {A} - \vec {v} _ {B} | t} \\ {\Rightarrow} & {t = \frac {| \vec {r} _ {B} - \vec {r} _ {A} |}{| \vec {v} _ {B} - \vec {v} _ {A} |}} \end{array}\tag{... (1}$
... (2)
Substituting (2) in (1) and rearrange to get (D). Hence, the correct answer is (D).

Since, $\frac{dv}{dt} = (g - cx^2)$
$\Rightarrow v \frac {d v}{d x} = g - c x ^ {2}$
$\begin{array}{r l} \Rightarrow & \int_ {v _ {0}} ^ {0} v d v = \int_ {0} ^ {x _ {m}} (g - c x ^ {2}) d x \\ \Rightarrow & - \frac {v _ {0} ^ {2}}{2} = g x _ {m} - \frac {c x _ {m} ^ {3}}{3} \\ \Rightarrow & \frac {c x _ {m} ^ {3}}{3} = g x _ {m} + \frac {v _ {0} ^ {2}}{2} \\ \Rightarrow & c = \frac {6 g x _ {m} + 3 v _ {0} ^ {2}}{2 x _ {m} ^ {3}} \end{array}$
Hence, the correct answer is (D).

$\begin{array}{r l} & s _ {n ^ {t h}} = u + \frac {a}{2} (2 n - 1) \\ & \Rightarrow s _ {n ^ {t h}} = 0 + \frac {a}{2} (2 n - 1) \\ & \Rightarrow s _ {n ^ {t h}} = \frac {a}{2} (2 n - 1) \\ & \Rightarrow s _ {1}: s _ {2}: s _ {3} \equiv 1: 3: 5 \end{array}$
Hence, the correct answer is (C).

$\vec{v} = \vec{u} +\vec{a} t$ $\Rightarrow \vec{v} = (6\hat{i} +8\hat{j}) + (0.8\hat{i} +0.6\hat{j})10$ $\Rightarrow \vec{v} = (6 + 8)\hat{i} +(8 + 6)\hat{j}$ $\Rightarrow \vec{v} = 14\hat{i} +14\hat{j}$ $\Rightarrow |\vec{v}| = 14\sqrt{2}\mathrm{ms}^{-2}$ Hence, the correct answer is (C).

Distance moved by the car is the area under v-t graph. So,
$\begin{array}{r l} & s _ {\mathrm{total}} = \frac {1}{2} (3) (3) + \frac {1}{2} (3) (3) \\ \Rightarrow & s _ {\mathrm{total}} = 9 \mathrm{m} \end{array}$

Hence, the correct answer is (C).

D

For OA, we have $a=\frac{5}{4}t$
$\Rightarrow d v = \frac {5}{4} t d t$
$\Rightarrow v _ {A} = \int d v = \frac {5}{4} \int_ {0} ^ {4} t d t = 1 0 \mathrm{ms} ^ {- 1}$
Otherwise, we can also observe that $\Delta v = \text{Area under a-t graph. So,}$
$\Delta v = v = 1 0 \mathrm{ms} ^ {- 1}$
For AB, we have
$v _ {B} = v _ {A} + a t$
$\Rightarrow v _ {B} = 1 0 + (5) (4) = 3 0 \mathrm{ms} ^ {- 1}$
Hence, the correct answer is (A).

For constant acceleration, assuming u = 0, then $v^{2} \propto x$ or $s \propto t^{2}$ Hence, the correct answer is (D).

$s =$ Area under the graph $\Rightarrow s = \frac{1}{2} (8)(2) + \frac{1}{2} (12)(3)$ $\Rightarrow s = 8 + 18$ $\Rightarrow s = 26\mathrm{m}$ Hence, the correct answer is (C).

Maximum acceleration is from C to D. So
$\begin{array}{r l} & a = \frac {6 0 - 2 0}{1 - 0 . 7 5} \\ \Rightarrow & a = \frac {4 0}{0 . 2 5} \\ \Rightarrow & a = 1 6 0 \mathrm{kmh} ^ {- 2} \\ & \text {Hence, the correct answer is (B).} \end{array}$

$\vec{v}_{\mathrm{man}} = \frac{v}{\sqrt{2}}\hat{i} +\frac{v}{\sqrt{2}}\hat{j}$ Let $\vec{v}_{\mathrm{wind}} = a\hat{i} +b\hat{j}$ $\Rightarrow \vec{v}_{\mathrm{wind / man}} = \left(a - \frac{v}{\sqrt{2}}\right)\hat{i} +\left(b - \frac{v}{\sqrt{2}}\right)\hat{j}$ $\Rightarrow \tan \theta = \frac{b - \frac{v}{\sqrt{2}}}{a - \frac{v}{\sqrt{2}}} = \tan (270)$ $\Rightarrow a - \frac{v}{\sqrt{2}} = 0$
$\begin{array}{r l} \Rightarrow & a = \frac {v}{\sqrt {2}} \\ \Rightarrow & \vec {v} _ {\mathrm{wind}} = \frac {v}{\sqrt {2}} \hat {i} + b \hat {j} \end{array}$
When the man doubles his speed
$\begin{array}{l l} & \vec {v} _ {\text {man}} ^ {\prime} = 2 \bigg (\frac {v}{\sqrt {2}} \hat {i} + \frac {v}{\sqrt {2}} \hat {j} \bigg) = \sqrt {2} (v \hat {i} + v \hat {j}) \\ \Rightarrow & \vec {v} _ {\text {wind / man}} ^ {\prime} = \bigg (\frac {v}{\sqrt {2}} - \sqrt {2} v \bigg) \hat {i} + (b - \sqrt {2} v) \hat {j} \\ \Rightarrow & \tan \theta^ {\prime} = \frac {b - \sqrt {2} v}{\frac {v}{\sqrt {2}} - \sqrt {2} v} = \frac {2 v - \sqrt {2} b}{v} \\ \text {But} & \theta^ {\prime} = 2 7 0 - \cot^ {- 1} (2) \\ \Rightarrow & \tan [ 2 7 0 - \cot^ {- 1} (2) ] = \frac {2 v - \sqrt {2} b}{v} \\ \Rightarrow & \cot [ \cot^ {- 1} (2) ] = \frac {2 v - \sqrt {2} b}{v} \\ \Rightarrow & 2 v = 2 v - \sqrt {2} b \\ \Rightarrow & b = 0 \\ \therefore & \vec {v} _ {\text {wind}} = \frac {v}{\sqrt {2}} \hat {i} \end{array}$
Hence, the correct answer is (B).

Distance between two cars leaving from station x is,
$d = \left(\frac {1}{6}\right) \times 6 0 = 1 0 \mathrm{km}$
Man meets the first car after time,
$t _ {1} = \frac {6 0}{6 0 + 6 0} = \frac {1}{2} \mathrm{h}$
He will meet the next car after time,
$t _ {2} = \frac {1 0}{6 0 + 6 0} = \frac {1}{1 2} \mathrm{h}$
In the remaining half an hour, number of cars he will meet again is, $n = \frac{1/2}{1/12} = 6$
So, total number of cars would be meet on route will be 7.
Hence, the correct answer is (B).

$h = \frac{1}{2} g t^2 \tag{... (1)}$
$\frac{h}{2} = \frac{1}{2} g (t - 1)^2 \tag{... (2)}$
Solving $(1)$ and $(2)$, we get:
\begin{aligned} t &= 2 \pm \sqrt{2}\text{ s} \\ \text{But } 2 - \sqrt{2} &< 1\text{ s} \\ \Rightarrow t &= 2 + \sqrt{2}\text{ s} \\ \Rightarrow t &= 3.41\text{ s} \end{aligned}
Hence, the correct answer is (B).

$\begin{array}{r l} & {\frac {d v}{A - B v} = d t} \\ & {\Rightarrow \int_ {0} ^ {v} \frac {d v}{A - B v} = \int_ {0} ^ {t} d t} \\ & {\Rightarrow \left. - \frac {1}{B} \log_ {e} (A - B v) \right| _ {0} ^ {v} = t} \\ & {\Rightarrow \log_ {e} \bigg (\frac {A - B v}{A} \bigg) = - B t} \\ & {\Rightarrow \frac {A - B v}{A} = e ^ {- B t}} \\ & {\Rightarrow v = \frac {A}{B} (1 - e ^ {- B t})} \end{array}$
Hence, the correct answer is (C).

Let the distance travelled by each car be s, then
$\begin{array}{r l} & v _ {1} - v _ {2} = v _ {0} \\ \Rightarrow & \sqrt {2 a _ {1} s} - \sqrt {2 a _ {2} s} = v _ {0} \text {and} \\ & t _ {2} - t _ {1} = t _ {0} \\ \Rightarrow & \sqrt {\frac {2 s}{a _ {2}}} - \sqrt {\frac {2 s}{a _ {1}}} = t _ {0} \\ \Rightarrow & \frac {v _ {0}}{t _ {0}} = \frac {\sqrt {a _ {1}} - \sqrt {a _ {2}}}{\frac {1}{\sqrt {a _ {2}}} - \frac {1}{\sqrt {a _ {1}}}} = \sqrt {a _ {1} a _ {2}} \end{array}$
Hence, the correct answer is (D).