Kinematics-1D

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.
$\begin{array}{c c c c c c c} & I & & I & & I \\ \hline A & t _ {1} & B & t _ {2} & C & t _ {3} & D \end{array}$
For AB
$l = \left(\frac {v _ {A} + v _ {B}}{2}\right) t _ {1}\tag{... (1}$
For BC
For the interval BC:
$l = \left(\frac{v_B + v_C}{2}\right) t_2 \tag{2}$
For the interval CD:
$l = \left(\frac{v_C + v_D}{2}\right) t_3 \tag{3}$
Combining equations (1), (2), and (3):
$\frac{1}{t_1} - \frac{1}{t_2} + \frac{1}{t_3} = \frac{1}{2l}(v_A + v_B - v_B - v_C + v_C + v_D)$
$\implies \frac{1}{t_1} - \frac{1}{t_2} + \frac{1}{t_3} = \frac{1}{2l}(v_A + v_D) \tag{4}$
For the entire interval AD:
$3l = \left(\frac{v_A + v_D}{2}\right) (t_1 + t_2 + t_3)$
$\implies \frac{1}{2l}(v_A + v_D) = \frac{3}{t_1 + t_2 + t_3} \tag{5}$
Finally, substituting (5) into (4):
$\frac{1}{t_1} - \frac{1}{t_2} + \frac{1}{t_3} = \frac{3}{t_1 + t_2 + t_3}$
$\implies k = 3$

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,
Here is the corrected and properly formatted LaTeX equation for the uniform speed interval shown in the highlighted section:
$400 = v t_2$
And for the subsequent accelerated interval:
$600 = \left(\frac{20 + v_2}{2}\right) t_3$

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
$\begin{array}{r l} & 2 0 0 = u (2) + \frac {1}{2} a (2) ^ {2} \\ \Rightarrow & u + a = 1 0 0 \\ & \begin{array}{c c c} & \longleftarrow 2 0 0 \mathrm{cm} & \longleftarrow 2 2 0 \mathrm{cm} \\ A & B & C \\ u, a & \longrightarrow 2 \mathrm{s} & \longrightarrow \\ & \longmapsto 6 \mathrm{s} & \longrightarrow \end{array} \end{array}\tag{... (1}$
For the interval AC:
$(200 + 220) = u(2 + 4) + \frac{1}{2}a(2 + 4)^2$
$\Rightarrow \quad 420 = 6u + 18a$
$\Rightarrow \quad u + 3a = 70$
Solving equations (1) and (2) gives:
$a = -15\text{ cm s}^{-2} \quad \text{and} \quad u = 115\text{ cm s}^{-1}$
If $v$ is the velocity at the end of the seventh second, then using $v = u + at$:
v = 115 + (-15)(7)
$\Rightarrow \quad v = 10\text{ cm s}^{-1}$

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
$\frac{dt}{d\alpha_1} = 0 \ \Rightarrow \ \frac{a}{v_1}\sec\alpha_1\tan\alpha_1 + \frac{b}{v_2}\sec\alpha_2\tan\alpha_2\left(\frac{d\alpha_2}{d\alpha_1}\right) = 0$
And for the derivative of constraint equation $(2)$ with respect to $\alpha_1$:
$a\sec^2\alpha_1 + b\sec^2\alpha_2\left(\frac{d\alpha_2}{d\alpha_1}\right) = 0 \ \Rightarrow \ \frac{d\alpha_2}{d\alpha_1} = -\frac{a\sec^2\alpha_1}{b\sec^2\alpha_2}$

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 $MC + CN = d = constant$
$\Rightarrow \quad a\tan {{\alpha }_{1}}+b\tan {{\alpha }_{2}}=\text{ constant }(\ldots (2)$
Taking derivative of (2) w.r.t. $\alpha_{1}$ on both sides, we get
$a \sec^2\alpha_1 + b \sec^2\alpha_2 \left(\frac{d\alpha_2}{d\alpha_1}\right) = 0 \quad \implies \quad \frac{d\alpha_2}{d\alpha_1} = -\frac{a \sec^2\alpha_1}{b \sec^2\alpha_2}$
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.

Initial relative velocity of approach of A and B is
$u _ {r} = (1 4 4 - 1 0 8) \mathrm{kmh} ^ {- 1} = 3 6 \mathrm{kmh} ^ {- 1} = 1 0 \mathrm{ms} ^ {- 1}$
The driver of train A applies brakes, so if a is the retardation, then retardation of A w.r.t. B is $a_{r} = -a$ . With this retardation he just manages to avoid the collision, so
$\begin{array}{r l} & v _ {r} ^ {2} - u _ {r} ^ {2} = 2 a _ {r} l _ {r} \\ \Rightarrow & O ^ {2} - (1 0) ^ {2} = 2 (- a) (1 0 0 0) \\ \Rightarrow & a = \frac {1 0 0}{2 0 0 0} = \frac {1}{2 0} \mathrm{ms} ^ {- 2} = 5 \mathrm{cms} ^ {- 2} \end{array}$
If this retardation is produced for a duration on t (say), then
$\begin{array}{r l} & v _ {r} = u _ {r} + a _ {r} t \\ \Rightarrow & 0 = (1 0) - \left(\frac {1}{2 0}\right) t \\ \Rightarrow & t = 2 0 0 \mathrm{s} \end{array}$

For convenience, let us take $\hat{i}$ along east, $\hat{j}$ along north and $\hat{k}$ vertically downwards.
$\vec {v} _ {b} = 8 \hat {i}, \vec {v} _ {s b} = 1 2 \hat {j} + 2 \hat {k}$
So, absolute velocity of submarine is
$\vec {v} _ {s} = \vec {v} _ {s b} + \vec {v} _ {b} = 8 \hat {i} + 1 2 \hat {j} + 2 \hat {k} \qquad \left\{\because \vec {v} _ {s b} = \vec {v} _ {s} - \vec {v} _ {b} \right\}$
$\begin{array}{r l} \Rightarrow & \vec {v} _ {h} = \vec {v} _ {h s} + \vec {v} _ {s} = 4 \hat {i} + 1 2 \hat {j} - 3 \hat {k} \qquad \left\{\because \vec {v} _ {h s} = \vec {v} _ {h} - \vec {v} _ {s} \right\} \\ \Rightarrow & | \vec {v} _ {h} | = \sqrt {(4) ^ {2} + (1 2) ^ {2} + (3) ^ {2}} = 1 3 \mathrm{ms} ^ {- 1} \end{array}$
Since, $\vec{v}_{hb} = \vec{v}_h - \vec{v}_b = -4\hat{i} +12\hat{j} -3\hat{k}$
$\Rightarrow \left| \vec {v} _ {h b} \right| = \sqrt {(- 4) ^ {2} + (1 2) ^ {2} + (- 3) ^ {2}} = 1 3 \mathrm{ms} ^ {- 1}$

(a) Distance travelled by the particle between t = 0 to 10 s is given by
$\begin{array}{l l} & s = \text { Area of } \Delta O A B = \frac {1}{2} \times 1 0 \times 1 2 = 6 0 \mathrm{m} \\ & \text { Average speed } = \frac {\text { total distance covered }}{\text { total time taken }} \\ \Rightarrow & \text { Average speed } = \frac {6 0}{1 0} = 6 \mathrm{ms} ^ {- 1} \end{array}$
(b) Acceleration of the particle during journey OA is given by
$v = u + at$ where $v = 12\mathrm{ms}^{-1}$ , $u = 0$ and $t = 5\mathrm{s}$
$\Rightarrow a = + 2. 4 \mathrm{ms} ^ {- 2}$
Similarly, acceleration of the particle during journey AB is given by
$v = u + at$ where $v = 0$ , $u = 12 \mathrm{~ms}^{-1}$ and $t = 5 \mathrm{~s}$
$\Rightarrow a = - 2. 4 \mathrm{ms} ^ {- 2}$
Velocity of the particle after 2 s from the start is given by
$v = u + a t = 0 + 2. 4 \times 2 = 4. 8 \mathrm{ms} ^ {- 1}$
So, distance covered by the particle between t = 2 to 5 s (in 3 s) is given by
$\begin{array}{r l} & s _ {1} = u t + \frac {1}{2} a t ^ {2} \\ \Rightarrow & s _ {1} = (4. 8) (3) + \frac {1}{2} (2. 4) (3) ^ {2} = 2 5. 2 \mathrm{m} \end{array}$
Distance covered by the particle from t = 5 s to 6 s (in 1 s) is given by
$\begin{array}{r l} & s _ {2} = u t + \frac {1}{2} a t ^ {2} \\ \Rightarrow & s _ {2} = (1 2) (1) + \frac {1}{2} (- 2. 4) (1) ^ {2} = 1 0. 8 \mathrm{m} \end{array}$
Total distance travelled from t = 2 s to 6 s is
$s = s _ {1} + s _ {2} = 2 5. 2 + 1 0. 8 = 3 6 \mathrm{m}$
Average speed in the interval t = 2 s to 6 s
$v _ {a v} = \frac {\mathrm{totaldistancecovered}}{\mathrm{totaltimetaken}} = \frac {3 6}{4} = 9 \mathrm{ms} ^ {- 1}$

$\frac{dx}{dt} = v_0 - \frac{v_0t}{5}$
$\Rightarrow x = v _ {0} t - \frac {v _ {0} t ^ {2}}{1 0}$
where x can be either +10 cm or -10 cm
For $x = +10 \, \mathrm{cm}$ , we have $10t - t^2 = 10$
$\Rightarrow \quad t = \left(\frac {1 0 \pm \sqrt {6 0}}{2}\right) s$
Hence we have
$t _ {1} = \left(\frac {1 0 - \sqrt {6 0}}{2}\right) \mathrm{s}$
{First Instant}
$\Rightarrow t _ {2} = \left(\frac {1 0 + \sqrt {6 0}}{2}\right) \mathrm{s}$
{Second Instant}
For $x = -10\mathrm{cm}$ , we have $10t - t^2 = -10$
$\Rightarrow t = \left(\frac {1 0 \pm \sqrt {1 4 0}}{2}\right) s$
$\Rightarrow t _ {3} = \left(\frac {1 0 + \sqrt {1 4 0}}{2}\right) \mathrm{s}$
{Third Instant}
However, we will not get fourth instant as time cannot be negative.
Time interval between the second and the third instant is $\Delta t = t_{3} - t_{2} \approx 2$ s

Initial velocity is $v_{i}=0$ and final velocity of the athlete
is $v_{f} = 36 \times \frac{5}{18} = 10 \, \mathrm{ms}^{-1}$ .
Since the particle is moving with uniform acceleration, so, we have
$v _ {a v} = \langle v \rangle = \frac {v _ {i} + v _ {f}}{2} = 5 \mathrm{ms} ^ {- 1}$