Kinematics-1D

A → (r) B → (s) C → (p) D → (q)

A → (s)
$\mathrm{C} \rightarrow (\mathrm{q})$
Acceleration is constant and is equal to acceleration due to gravity, which is acting vertically downwards i.e. in the positive direction, so (p) represents the a-t graph for the motion.
Since $v = u + at$ and $s = \Delta y = ut + \frac{1}{2}gt^{2}$

Before collision, v = gt {∵ u = 0}
So, before collision velocity is increasing linearly with time and is increasing. However, after collision velocity decreases with time linearly and is negative as in (r).
$y = \frac {1}{2} g t ^ {2}$
tive, whereas after collision, the displacement is $y = -v_{0}t + \frac{1}{2}gt^{2}$ and is positive (a parabola opening
(upwards). Here is the tricky part, as we have taken the origin at the point from where the ball is dropped and in both the cases of upward and downward motion, the displacement is downwards and hence displacement is positive in both the cases.
The distance-time graph is positive and always increasing.

A → (p, r) B → (p, r) C → (p, q) D → (p, q)

A → (p, q)
$\mathrm{B} \rightarrow (\mathrm{p}, \mathrm{q})$
$\mathrm{D} \rightarrow (\mathrm{p}, \mathrm{q})$
With constant positive acceleration, speed increases, when velocity is positive and speed decreases, when velocity is negative.
Similarly, with constant negative acceleration speed increases, when velocity is negative and speed decreases, when velocity is positive.

A → (q)
B → (r)
C → (s)
D → (p)
Solve using relative velocity diagram.

A → (p, r) B → (q, r) C → (q) D → (q, s)

A → (q)
Particle will change the direction of motion
$\mathrm{when} \frac {d x}{d t} = 0$
$\Rightarrow - 2 0 + 1 0 t = 0$
$\Rightarrow t = 2 \mathrm{s}$

Let velocities at $A$, $B$, $C$, and $D$ be $v_A$, $v_B$, $v_C$, and $v_D$ respectively.
For segment $AB$:
$l = \left(\frac{v_A + v_B}{2}\right) t_1$
For segment $BC$:
$l = \left(\frac{v_B + v_C}{2}\right) t_2 \tag{... (2)}$
For segment $CD$:
$l = \left(\frac{v_C + v_D}{2}\right) t_3 \tag{... (3)}$
From $(1)$, $(2)$, and $(3)$, we have:
$\frac{1}{t_1} - \frac{1}{t_2} + \frac{1}{t_3} = \frac{1}{2l} (v_A + v_B - v_B - v_C + v_C + v_D)$
$\Rightarrow \frac{1}{t_1} - \frac{1}{t_2} + \frac{1}{t_3} = \frac{1}{2l} (v_A + v_D) \tag{... (4)}$
For segment $AD$:
$3l = \left(\frac{v_A + v_D}{2}\right) (t_1 + t_2 + t_3)$
$\Rightarrow \frac{1}{2l} (v_A + v_D) = \frac{3}{t_1 + t_2 + t_3} \tag{... (5)}$
So, from $(4)$ and $(5)$, we get:
$\frac{1}{t_1} - \frac{1}{t_2} + \frac{1}{t_3} = \frac{3}{t_1 + t_2 + t_3}$
$\Rightarrow k = 3$

Since time to accelerate from zero to $60 \, kmh^{-1}$ is equal to the time taken to decelerate from $60 \, kmh^{-1}$ to zero. Hence, we can say both OA and BC are identical intervals. So, total distance travelled from O to A to B to C is
$l = l _ {1} + l _ {2} + l _ {1} = \frac {1}{2} v _ {\mathrm{max}} t + (v _ {\mathrm{max}}) 8 t + \frac {1}{2} v _ {\mathrm{max}} t$
Since,
Average Speed= $\frac{Total Distance Travelled}{Total Time Taken}$
$\Rightarrow v _ {a v} = \frac {l _ {1} + l _ {2} + l _ {1}}{t + 8 t + t}$

$\begin{array}{l l} \Rightarrow & v _ {a v} = \frac {\frac {1}{2} v _ {\max} t + (v _ {\max}) 8 t + \frac {1}{2} v _ {\max} t}{1 0 t} \\ \Rightarrow & v _ {a v} = \frac {9}{1 0} v _ {\max} = \frac {9}{1 0} (6 0 \mathrm{kmh} ^ {- 1}) = 5 4 \mathrm{kmh} ^ {- 1} \\ \Rightarrow & v _ {a v} = 5 4 \times \frac {5}{1 8} \mathrm{ms} ^ {- 1} = 1 5 \mathrm{ms} ^ {- 1} \end{array}$
$\begin{array}{l l} 3. & u = 2 7 \mathrm{ms} ^ {- 1} \text {at} t _ {0} = 0 \mathrm{s}. \text {Since} \\ & d v = a d t \\ & \Rightarrow \int_ {2 7} ^ {v} d v = \int_ {0} ^ {t} - 6 t d t \\ & \Rightarrow v = (2 7 - 3 t ^ {2}) \mathrm{ms} ^ {- 1} \\ & \text {At} v = 0, \text {from equation ... (1)} \\ & 0 = 2 7 - 3 t ^ {2} \\ & \Rightarrow t = 3 \mathrm{s} \\ & \text {So, it will stop at} t = 3 \mathrm{s} \\ & \text {Since} d x = v d t \\ & \Rightarrow \int_ {0} ^ {x} d x = \int_ {0} ^ {3} (2 7 - 3 t ^ {2}) d t \\ & \Rightarrow x = (2 7 t - t ^ {3}) | _ {0} ^ {3} = 2 7 (3) - (3) ^ {3} = 5 4 \mathrm{m} \end{array}\tag{... (1}$

54

$x_{\mathrm{total}} = (0.5 + 1.5 + 1.5 + 2.5) = 6 \mathrm{~m}$

$t = (2 + 4) = 6 \mathrm{s}$
$\left(v _ {s p}\right) _ {a v g} = \frac {x _ {\mathrm{total}}}{t} = \frac {6}{6} = 1 \mathrm{ms} ^ {- 1}$

$\begin{array}{l l} \text {(a)} & 8 0 = \left(\frac {u + 1 8}{2}\right) 1 0 \\ & \Rightarrow \quad 1 6 = u + 1 8 \\ & \xrightarrow {\text {1 m}} \\ & t = 1 \xrightarrow {\text {u = 2 ms} ^ {- 1}} t = 0 \xrightarrow {\text {a}} \xrightarrow {\oplus} \\ & \xrightarrow {\text {1 m}} \xrightarrow {\text {80 m}} \\ & \Rightarrow \quad u = - 2 \mathrm{ms} ^ {- 1} \\ \text {(b)} & 1 8 = - 2 + a (1 0) \\ & \Rightarrow \quad a = 2 \mathrm{ms} ^ {- 2} \\ \text {(c)} & v ^ {2} - u ^ {2} = 2 a s \\ & \Rightarrow \quad 0 ^ {2} - (2) ^ {2} = 2 (- 2) s \\ & \Rightarrow \quad s = 1 \mathrm{m} \\ & \qquad \qquad \qquad \qquad \qquad 0 = - 2 + (2) t \\ & \Rightarrow \quad t = 1 \mathrm{s} \\ \text {(d)} & s _ {\text {total}} = 8 0 + | - 1 | + | 1 | = 8 2 \mathrm{m} \end{array}$

For normal driver, the car moves a distance of $x_{0}=vt=44(0.75)=33\ m$ before he or she reacts and decelerates the car. The stopping distance can be obtained using
$\begin{array}{r l} & v ^ {2} - u ^ {2} = 2 a (x - x _ {0}), \\ \text {where} & x = ?, x _ {0} = 3 3 \mathrm{m}, a = - 2 \mathrm{ms} ^ {- 2}, a = 4 4 \mathrm{ms} ^ {- 1} \\ \Rightarrow & 0 ^ {2} - 4 4 ^ {2} = 2 (- 2) (x - 3 3) \\ \Rightarrow & x = 5 1 7 \mathrm{m} \end{array}$
For a drunk driver, the car moves a distance of $x_{0}^{\prime}=vt=44(3)=132\ m$ before he or she reacts and decelerates the car. The stopping distance can be obtained again by using
$v ^ {2} - u ^ {2} = 2 a \left(x ^ {\prime} - x _ {0} ^ {\prime}\right)$
$\begin{array}{r l} \Rightarrow & 0 ^ {2} - 4 4 ^ {2} = 2 (- 2) (x ^ {\prime} - 1 3 2) \\ & x ^ {\prime} = 6 1 6 \mathrm{m} \end{array}$

(a) Let the bus overtake the car after a distance x. When the two meet, the time taken (say t) is the same. So, for bus,
$x = 0 + \frac {1}{2} 5 t ^ {2}$
$\left\{\because s = u t + \frac {1}{2} a t ^ {2} \right\}$
$\begin{array}{r l} & {\mathrm{Forcar,} x = 5 0 t} \\ & {\Rightarrow \quad \frac {5}{2} t ^ {2} = 5 0 t} \\ & {\Rightarrow \quad t = 2 0 \mathrm{s}} \\ & {\Rightarrow \quad x = (5 0) (2 0) = 1 0 0 0 \mathrm{m}} \end{array}$
(b) Let the bus be moving with a velocity v.
So, $v^{2}-0=2(5)(1000)$ $\left\{\because v^{2}-u^{2}=2as\right\}$ $\Rightarrow \quad v = 100 \, ms^{-1}$

Velocity of truck is $V = 72 \, kmh^{-1} = 20 \, ms^{-1}$ Time to overtake the truck, 0.6 km down the road is
$t = \frac {d}{V} = \frac {0 . 6 \times 1 0 ^ {3}}{2 0} = 3 0 \mathrm{s}$
If u be the initial velocity of the car, then
$6 0 0 = u (3 0) + \frac {1}{2} (1) (3 0) ^ {2}$
Solving it we have $u = 5 \, ms^{-1}$
Velocity of the car at the instant of overtaking is
$v = 5 + (1) (3 0) = 3 5 \mathrm{ms} ^ {- 1}$
$\left\{\because v = u + a t \right\}$
The relative velocity of car with respect to truck is
$v _ {r} = 3 5 - 2 0 = 1 5 \mathrm{ms} ^ {- 1}$
The distance between them at t = 50 s, i.e., 20 s after overtaking is
$s = (1 5) (2 0) = 3 0 0 \mathrm{m}$

Since $s=\left(\frac{u+v}{2}\right)t$ , so for decelerated interval,
$2 0 0 = \left(\frac {2 0 + v}{2}\right) t _ {1}\tag{... (1}$
For uniform interval,
$400= v t _ {2}$ ....(2)
and again for accelerated interval, we have
$6 0 0 = \left(\frac {2 0 + v}{2}\right) t _ {3}$ (3)

Given $t_{2}=t_{1}+t_{3}$
$\begin{array}{r l} & {\Rightarrow \frac {4 0 0}{v} = \frac {4 0 0}{2 0 + v} + \frac {1 2 0 0}{2 0 + v}} \\ & {\Rightarrow \frac {4 0 0}{v} = \frac {1 6 0 0}{2 0 + v}} \\ & {\Rightarrow 2 0 + v = 4 v} \\ & {\Rightarrow v = \frac {2 0}{3} \mathrm{ms} ^ {- 1}} \\ & {\Rightarrow t _ {1} = \frac {4 0 0}{2 0 + \frac {2 0}{3}} = \frac {1 2 0 0}{8 0} \mathrm{s} = 1 5 \mathrm{s}} \\ & {\Rightarrow t _ {2} = \frac {4 0 0}{v} = \frac {4 0 0}{\frac {2 0}{3}} = \frac {1 2 0 0}{2 0} = 6 0 \mathrm{s}} \\ & {\Rightarrow t _ {3} = \frac {1 2 0 0}{2 0 + v} = \frac {1 2 0 0}{2 0 + \frac {2 0}{3}} = \frac {3 6 0 0}{8 0} \mathrm{s} = 4 5 \mathrm{s}} \\ & {\Rightarrow t = (1 5 + 6 0 + 4 5) \mathrm{s} = 1 2 0 \mathrm{s}} \\ & {\Rightarrow t = 2 \mathrm{minute}} \end{array}$

Let the truck overtake the car at time t = 0 and the car overtake the truck at time t.
For truck, s = 10t, and for car, $s = \frac{1}{2}2t^{2}$
$\begin{array}{r l} & 1 0 t = \frac {1}{2} (2) t ^ {2} \\ \Rightarrow & t = 1 0 \mathrm{s} \\ \Rightarrow & s = \frac {1}{2} (2) (1 0) ^ {2} = 1 0 0 \mathrm{m} \\ \text { and } & v = (2) (1 0) = 2 0 \mathrm{ms} ^ {- 1} \end{array}$

Let u be the initial velocity of the bullet.
According to the problem $v=\frac{u}{2}$ , when s=3 cm
$\begin{array}{r l} \Rightarrow & \left(\frac {u}{2}\right) ^ {2} - u ^ {2} = 2 a (3) \\ \Rightarrow & a = \frac {u ^ {2}}{8} \end{array}\tag{... (1}$
Let the bullet travels a distance x cm before coming to rest so,
$\begin{array}{r l} & 0 - u ^ {2} = 2 a x \\ \Rightarrow & u ^ {2} = 2 \left(\frac {u ^ {2}}{8}\right) x \\ \Rightarrow & x = 4 \mathrm{cm} \end{array}$
So the bullet will move 4-3=1 cm further before coming to rest

Let the particle start with initial velocity u and a uniform acceleration a from point A.
For AB
$2 0 0 = u (2) + \frac {1}{2} a (2) ^ {2}$
$\Rightarrow$ $u + a = 1 0 0$
${c c c}$ $\longleftarrow 2 0 0 \mathrm{cm}$
$\longleftarrow 2 2 0 \mathrm{cm}$ A & B & C $u, a $
$\longrightarrow 2 \mathrm{s}$
$\longrightarrow$ $\longmapsto 6 \mathrm{s}$ $\longrightarrow$
For AC
$r l (2 0 0 + 2 2 0) = u (2 + 4) + \frac {1}{2} a (2 + 4) ^ {2}$
$\Rightarrow$
$4 2 0 = 6 u + 1 8 a$
$\Rightarrow$
u + 3 a = 7 0 Solving 1) and 2), we get $a = - 1 5 \mathrm{cms} ^ {- 2} \text {and} u = 1 1 5 \mathrm{cms} ^ {- 1}$....(2)
If v be the velocity at the end of seventh second from the start, then using
$r l v = u + a t, \mathrm{weget}$
$\Rightarrow v = 1 1 5 + (- 1 5) (7)$
$ \Rightarrow v = 1 0 \mathrm{cms} ^ {- 1}$

For OB
$\begin{array}{c} a = \frac {v _ {2} - v _ {1}}{\left(t _ {2} + t _ {1}\right) - 0} = \frac {v _ {2} - v _ {1}}{t _ {2} + t _ {1}} \\ \longmapsto \begin{array}{c c c c c c c c} \longmapsto & t _ {1} & \longmapsto & t _ {2} & \longmapsto & \longmapsto & t _ {3} & \longmapsto \\ O & v _ {1} & A & v _ {2} & B & v _ {3} & C \end{array} \end{array}\tag{... (1}$
For AC
$a = \frac {v _ {3} - v _ {2}}{(t _ {3} + t _ {2} + t _ {1}) - t _ {1}} = \frac {v _ {3} - v _ {2}}{t _ {3} + t _ {2}}\tag{... (2}$
From (1) and (2), equating the values of a, we get
$\left(\frac {v _ {1} - v _ {2}}{v _ {2} - v _ {3}}\right) \left(\frac {t _ {3} + t _ {2}}{t _ {2} + t _ {1}}\right) = 1$

For $0 \leq x < 1000 \, m$ , the initial condition is v = 0 at x = 0
Since $v dv = adx$
$\Rightarrow \int_ {0} ^ {v} v d v = \int_ {0} ^ {x} 6 d x \quad \left\{\because a = 6 \mathrm{ms} ^ {- 2} \text { from graph } \right\}$
$\Rightarrow \frac {v ^ {2}}{2} = 6 x$
$\Rightarrow v = 2 \sqrt {3 x}\tag{... (1}$
When $x = 1000\mathrm{m}$ , we get
$v = 2 \sqrt {3 0 0 0} = 2 0 \sqrt {3 0} \mathrm{ms} ^ {- 1}$
For $1000\mathrm{m} < x \leq x_0$ , the initial condition is
$v = 2 0 \sqrt {3 0} \mathrm{ms} ^ {- 1} \mathrm{at} x = 1 0 0 0 \mathrm{m}$
Since $v dv = adx$
$\Rightarrow \int_ {2 0 \sqrt {3 0}} ^ {v} v d v = \int_ {1 0 0 0} ^ {x} - 4 d x$
$\Rightarrow \quad \frac {v ^ {2}}{2} \bigg | _ {2 0 \sqrt {3 0}} ^ {v} = - 4 x \bigg | _ {1 0 0 0} ^ {x}$
$\Rightarrow \quad \frac {v ^ {2}}{2} - \frac {1 2 0 0 0}{2} = - 4 (x - 1 0 0 0)$
$\Rightarrow v ^ {2} - 1 2 0 0 0 = - 8 x + 8 0 0 0$
$\Rightarrow v ^ {2} = 2 0 0 0 0 - 8 x$
$\Rightarrow v = \left(\sqrt {2 0 0 0 0 - 8 x}\right) \mathrm{ms} ^ {- 1}$
When $v = 0$ , at $x = x_0$ , then
$\Rightarrow 0 = \sqrt {2 0 0 0 0 - 8 x _ {0}}$
$\Rightarrow x _ {0} = 2 5 0 0 \mathrm{m}$

The velocity v of the balloon when it has risen to a height H is given by
$v ^ {2} = 0 + 2 \left(\frac {g}{8}\right) H$
taking upward direction positive
$\Rightarrow v = \sqrt {\frac {g H}{4}} = \frac {\sqrt {g H}}{2} \mathrm{msec} ^ {- 1}$
This will also be the velocity of the stone in upward direction when it is dropped.
Taking upward direction positive for the stone,
$- H = \left(\frac {\sqrt {g H}}{2}\right) t - \frac {1}{2} g t ^ {2} \quad \left\{\because g = - g \text {and} h = - H \right\}$
$\Rightarrow g t ^ {2} - \sqrt {g H} t - 2 H = 0$
$\Rightarrow t = \frac {\sqrt {g H} \pm \sqrt {g H + 8 g H}}{2 g}$
$\Rightarrow t = \frac {\sqrt {g H} \pm 3 \sqrt {g H}}{2 g}$
$\Rightarrow t = 2 \sqrt {\frac {H}{g}}$
{taking positive value}
$\Rightarrow x = 2$

For the first kilometre of the journey,
$v ^ {2} - u ^ {2} = 2 a x$
$\Rightarrow 1 0 0 - 4 = 2 a (1 0 0 0)$
$\Rightarrow a = 0. 0 4 8 \mathrm{ms} ^ {- 2}$
For the second kilometre, again applying
$v ^ {2} - u ^ {2} = 2 a s$
where $v = ?$ , $u = 10 \, \mathrm{ms}^{-1}$ , $a = 0.048 \, \mathrm{ms}^{-2}$ and $s = 1000 \, \mathrm{m}$
$\Rightarrow v = 1 4 \mathrm{ms} ^ {- 1}$
For the whole journey, $v_{0}=2\ ms^{-1}$ , $v=14\ ms^{-1}$ and $0.048\ ms^{-2}$ . So,
v = k + a t
$\Rightarrow 1 4 = 2 + 0. 0 4 8 t$
$\Rightarrow t = 2 5 0 \mathrm{s}$

$t = \frac {A C}{v _ {1}} + \frac {C B}{v _ {2}}$
where $AC = a \sec \alpha_{1}$ and $CB = b \sec \alpha_{2}$
$\Rightarrow t = \frac {a \sec \alpha_ {1}}{v _ {1}} + \frac {b \sec \alpha_ {2}}{v _ {2}}$
For $t_1$ to be MINIMUM, we have
$\begin{array}{l} \frac {d t}{d \alpha_ {1}} = 0 \\ \Rightarrow \quad \frac {a}{v _ {1}} \sec \alpha_ {1} \tan \alpha_ {1} + \frac {b}{v _ {2}} \sec \alpha_ {2} \tan \alpha_ {2} \left(\frac {d \alpha_ {2}}{d \alpha_ {1}}\right) = 0 \end{array}\tag{... (1}$

Please note that the question itself asks us to adjust the values of $\alpha_{1}$ and $\alpha_{2}$ . So both $\alpha_{1}$ and $\alpha_{2}$ are variables. Now, since $MC + CN = d = constant$
$\Rightarrow \quad a \tan \alpha_ {1} + b \tan \alpha_ {2} =$
constant...(2)
Taking derivative of (2) w.r.t. $\alpha_{1}$ on both sides, we get
$\begin{array}{r l} & a \sec^ {2} \alpha_ {1} + b \sec^ {2} \alpha \left(\frac {d \alpha_ {2}}{d \alpha_ {1}}\right) = 0 \\ \Rightarrow & \frac {d \alpha_ {2}}{d \alpha_ {1}} = - \frac {a \sec^ {2} \alpha_ {1}}{b \sec^ {2} \alpha_ {2}} \end{array}\tag{... (3}$
Substituting (3) in (1), we get
$\begin{array}{r l} & \frac {\not {a}}{v _ {1}} \sec \alpha_ {1} \tan \alpha_ {1} + \frac {\not {b}}{v _ {2}} \sec \alpha_ {2} \tan \alpha_ {2} \left(- \frac {\not {a}}{\not {b}} \frac {\sec^ {2} \alpha_ {1}}{\sec^ {2} \alpha_ {2}}\right) \\ & = 0 \\ & \frac {\tan \alpha_ {1}}{v _ {1}} = \frac {\tan \alpha_ {2} \sec \alpha_ {1}}{v _ {2} \sec \alpha_ {2}} \\ & \frac {1}{v _ {1}} = \frac {\sin \alpha_ {1}}{\cos \alpha_ {1}} = \frac {1}{v _ {2}} \left(\frac {\sin \alpha_ {2}}{\cos \alpha_ {2}}\right) \left(- \frac {\cos \alpha_ {2}}{\cos \alpha_ {1}}\right) \\ & \Rightarrow \quad \frac {\sin \alpha_ {1}}{\sin \alpha_ {2}} = \frac {v _ {1}}{v _ {2}} \\ & \text {Since,} v _ {1} = 3 \sqrt {3} \mathrm{ms} ^ {- 1}, \alpha_ {1} = 3 0 ^ {\circ} \text {and} \alpha_ {2} = 6 0 ^ {\circ} \\ & \Rightarrow \quad \frac {\sin (3 0 ^ {\circ})}{\sin (6 0 ^ {\circ})} = \frac {3 \sqrt {3}}{v _ {2}} \\ & \Rightarrow v _ {2} = 9 \mathrm{ms} ^ {- 1} \end{array}$

$\begin{array} v _ {m} = \text { absolute velocity of man }, \\ & v _ {m e} = \text { velocity of man w.r.t. escalator }, \\ & v _ {e} = \text { velocity of escalator } \end{array}$
If the length of the elevator by $\ell$ , then from the given condition,
$\begin{array}{l} v _ {m} = \frac {l}{9 0} \mathrm{ms} ^ {- 1} \\ v _ {e} = \frac {l}{6 0} \mathrm{ms} ^ {- 1} \end{array}$
Time taken by the person to walk up in the moving escalator is,
$t = \frac {l}{v _ {e} + v _ {m}} = \frac {l}{\frac {l}{9 0} + \frac {l}{6 0}} = \frac {9 0 \times 6 0}{9 0 + 6 0} = 3 6 \mathrm{s}$
Here we observe that the time t does not depend on $\ell$ the length of escalator.