Kinematics-1D

$t=\frac{d}{u-u\cos60^{\circ}}$ $=\frac{d}{u-\frac{u}{2}}=\frac{d}{\frac{u}{2}}=\frac{2d}{u}$ Alternative method:- $t=\frac{d}{u\cos60^{\circ}}$ $t=\frac{2d}{u}$

speed of man $(\mathrm{v_m}) = \frac{\mathrm{d}}{45}$ speed of escalator $(\mathrm{v_s}) = \frac{\mathrm{d}}{60}$ $t = \frac{d}{\mathbf{v}_m + \mathbf{v}_s} = \frac{d}{\frac{d}{45} + \frac{d}{60}} = \frac{60\times 45}{105} = 25.71\sec$

cycle bus $\rightarrow$ 2m/s $^{2}$ For cycle 96 + s = 20 × t
for bus s = $\frac{1}{2}$ at $^{2}$ = $\frac{1}{2}$ × 2 × t $^{2}$ = t $^{2}$ ∴ 96 + t $^{2}$ = 20 t $\Rightarrow$ t $^{2}$ - 20t + 96 = 0
(t - 8) (t - 12) = 0 ∴ t = 8 s or 12 s
cyclist over takes bus in 8 s
Note: At t = 12s, bus overtakes cyclist.

$\frac{10}{\mathrm{v}_{\mathrm{mr}}} + \frac{10}{\mathrm{v}_{\mathrm{mr}}} = 2 \Rightarrow \mathrm{v}_{\mathrm{mr}} = 10 \mathrm{~km} / \mathrm{hr}$ . $\Rightarrow t = \frac{10}{\mathrm{v}_{\mathrm{mr}} + \mathrm{v}_{\mathrm{r}}} + \frac{10}{\mathrm{v}_{\mathrm{mr}} - \mathrm{v}_{\mathrm{r}}}$ $= \frac{10}{10 + 5} + \frac{10}{10 - 5} = \frac{10}{15} + \frac{10}{5} = \frac{8}{3} \mathrm{hr}$ $\Rightarrow t = \frac{8}{3} \times 60 = 160 \mathrm{~min}$ .

The velocity upstream is $(6 - 2)$ km/hr and downstream is $(6 + 2)$ km/hr.
∴ Total time taken = $\frac{2}{4} + \frac{2}{8} = \frac{6}{8} = \frac{3}{4}$ hr
= 45 minutes

Distance will remain the same as before because acceleration of both the bodies is same and initial velocity is zero.

Let us assume they are meeting at x distance from ground
$8 0 - \mathrm{x} = 2 0 \mathrm{t} + \frac {1}{2} \mathrm{gt} ^ {2}$
$\mathrm{x} = 2 0 \mathrm{t} - \frac {1}{2} \mathrm{gt} ^ {2}$
... (ii)
From eq. (i) and (ii)
80 = 40 t
t = 2 s
x = 20 m from ground.

Total time taken by the first walnut to reach at ground :-
$\mathrm{h} = \frac {1}{2} \mathrm{gt} ^ {2}$
$2 0 = \frac {1}{2} \times 1 0 \times t ^ {2}$
$t = 2 \mathrm{sec}.$
time taken by it to cover a distance of 5 m :
$5 = \frac {1}{2} \mathrm{gt} ^ {2} \Rightarrow t = 1 \sec .$
So, For 2 $^{nd}$ walnut :
$S = U t + \frac {1}{2} a t ^ {2}$
$2 0 = \mathrm{U} \times 1 + \frac {1}{2} \times 1 0 \times (1) ^ {2}$
$\mathrm{U} = 1 5 \mathrm{m/s}$

Let the two bodies A and B respectively meet at a time t, at a height $\frac{H}{2}$ from the ground.

Using S = ut + $\frac{1}{2}$ at $^{2}$
For a body A, $u = v_{0}$ , a = -g, $S = \frac{H}{2}$
$\therefore \frac {\mathrm{H}}{2} = \mathrm{v} _ {0} \mathrm{t} - \frac {1}{2} \mathrm{gt} ^ {2}$
... (i)
For a body B, u = 0, a = +g, S = $\frac{H}{2}$
$\therefore \frac {\mathrm{H}}{2} = \frac {1}{2} \mathrm{gt} ^ {2}$
Equating equations (i) and (ii), we get
$\mathrm {v_ {0} t- \frac {1}{2} g t^ {2} = \frac {1}{2} g t^ {2}}$
$\mathrm{v} _ {0} \mathrm{t} = \mathrm{gt} ^ {2} \text {or} \mathrm{t} = \frac {\mathrm{v} _ {0}}{\mathrm{g}}$
Substituting the value of $t$ is equation (i), we get
$\frac {\mathrm{H}}{2} = \mathrm{v} _ {0} \times \left(\frac {\mathrm{v} _ {0}}{\mathrm{g}}\right) - \frac {1}{2} \mathrm{g} \left(\frac {\mathrm{v} _ {0}}{\mathrm{g}}\right) ^ {2} = \frac {\mathrm{v} _ {0} ^ {2}}{\mathrm{g}} - \frac {1}{2} \frac {\mathrm{v} _ {0} ^ {2}}{\mathrm{g}}$
$\frac {\mathrm{H}}{2} = \frac {1}{2} \frac {\mathrm{v} _ {0} ^ {2}}{\mathrm{g}} \text {or} \mathrm{v} _ {0} ^ {2} = \mathrm{gH}$
$\mathbf {v} _ {0} = \sqrt {\mathrm{gH}}$
Alternate solution: -
$\mathbf {U} _ {\mathrm{AB}} = \mathbf {U} _ {\mathrm{A}} - \mathbf {U} _ {\mathrm{B}} = \mathbf {v} _ {0} - 0 = \mathbf {v} _ {0}$
$a _ {A B} = a _ {A} - a _ {B} = - g - (- g) = 0, S _ {A B} = H$
$\therefore \mathrm{t} = \frac {\mathrm{S} _ {\mathrm{AB}}}{\mathrm{U} _ {\mathrm{AB}}} = \frac {\mathrm{H}}{\mathrm{v} _ {0}}, \text {for body B}: - \mathrm{S} = 1 / 2 \mathrm{gt} ^ {2}$
$\Rightarrow \quad \frac {\mathrm{H}}{2} = \frac {1}{2} \mathrm{g} \times \frac {\mathrm{H} ^ {2}}{\mathrm{v} _ {0} ^ {2}} \Rightarrow \mathrm{v} _ {0} = \sqrt {\mathrm{gH}}$

Effective acceleration in ascending lift = (g + a)
$t = \sqrt {\frac {2 h}{g + a}} = \sqrt {\frac {2 \times 1 9}{3 2 + 6}}$
$\mathrm{t} = \sqrt {\frac {2 \times 1 9}{3 8}} = 1 \sec$

$\mathbf {v} _ {\mathrm{R}} ^ {2} + 8 ^ {2} = 1 0 ^ {2}$
$\mathrm{v} _ {\mathrm{R}} = 6 \mathrm{Km/hr}$

$\vec {\mathrm{v}} _ {\mathrm{bT}} = - 4 0 \hat {\mathrm{j}} - 4 0 \hat {\mathrm{i}}$
$\left| \vec {\mathrm{v}} _ {\mathrm{bT}} \right| = \sqrt {4 0 ^ {2} + 4 0 ^ {2}} = 4 0 \sqrt {2} (\text {south - west})$
$\tan \theta = 1; \theta = 4 5 ^ {\circ}$

$\vec {\bf {V}} _ {\mathrm{bT}} = \vec {\bf {V}} _ {\mathrm{b}} - \vec {\bf {V}} _ {\mathrm{T}}$
$\vec {\mathrm{v}} _ {\mathrm{bT}} = 4 0 \hat {\mathrm{j}} - (- 3 0 \hat {\mathrm{i}})$
$\vec {\mathrm{v}} _ {\mathrm{bT}} = (3 0 \hat {\mathrm{i}} + 4 0 \hat {\mathrm{j}}) \mathrm{km/h}$
$\left| \vec {\mathrm{v}} _ {\mathrm{bT}} \right| = \sqrt {(3 0) ^ {2} + (4 0) ^ {2}} = 5 0 \mathrm{km/h}$
$\tan \theta = \frac {4 0}{3 0} = \frac {4}{3}$
$\theta = 53^{\circ}$
$53^{\circ}$ North of east

$\vec{\mathrm{v}}_{\mathrm{BW}} = \vec{\mathrm{v}}_{\mathrm{BG}} - \vec{\mathrm{v}}_{\mathrm{WG}} = 9\hat{\mathrm{i}} +12\hat{\mathrm{j}}$

Here,
$\vec {\mathrm{r}} _ {1} (\mathrm{t}) = 3 \mathrm{t} \hat {\mathrm{i}} + 4 \mathrm{t} ^ {2} \hat {\mathrm{j}}$
$\vec {\mathrm{r}} _ {2} (\mathrm{t}) = 4 \mathrm{t} ^ {2} \hat {\mathrm{i}} + 3 \mathrm{t} \hat {\mathrm{j}}$
Velocity,
$\vec {v} _ {1} (t) = \frac {d \vec {r} _ {1}}{d t} = \frac {d}{d t} \left(3 t \hat {i} + 4 t ^ {2} \hat {j}\right) = 3 \hat {i} + 8 t \hat {j}$
$\vec {\mathrm{v}} _ {2} (t) = \frac {\mathrm{d} \vec {\mathrm{r}} _ {2}}{\mathrm{dt}} = \frac {\mathrm{d}}{\mathrm{dt}} \left(4 t ^ {2} \hat {\mathrm{i}} + 3 t \hat {\mathrm{j}}\right) = 8 t \hat {\mathrm{i}} + 3 \hat {\mathrm{j}}$
The relative speed of particle 1 with respect to particle 2 is: $\vec{\mathbf{v}}_{12} = \vec{\mathbf{v}}_1 - \vec{\mathbf{v}}_2$
$= (3 \hat {\mathrm{i}} + 8 \mathrm{t} \hat {\mathrm{j}}) - (8 \mathrm{t} \hat {\mathrm{i}} + 3 \hat {\mathrm{j}}) = (3 - 8 \mathrm{t}) \hat {\mathrm{i}} + (8 \mathrm{t} - 3) \hat {\mathrm{j}}$
$\mathrm{At} t = 2 \mathrm{s},$
$\vec {\mathrm{v}} _ {1 2} = (3 - 1 6) \hat {\mathrm{i}} + (1 6 - 3) \hat {\mathrm{j}} = - 1 3 \hat {\mathrm{i}} + 1 3 \hat {\mathrm{j}}$
$\left| \vec {\mathrm{v}} _ {1 2} \right| = \sqrt {(1 3) ^ {2} + (1 3) ^ {2}} = 1 3 \sqrt {2} \mathrm{m/s}$

$\vec {\mathbf {v}} _ {\mathrm{RM}} = \vec {\mathbf {v}} _ {\mathrm{R}} - \vec {\mathbf {v}} _ {\mathrm{M}} = - 3 \hat {\mathbf {j}} - 3 \hat {\mathbf {i}}$
$\vec {\mathrm{v}} _ {\mathrm{RM}} = \sqrt {3 ^ {2} + 3 ^ {2}} = 4. 2 \mathrm{m/s}$

$\vec{V}_{M}=5\hat{i}$
$\vec {V} _ {\mathrm{R}} = - 1 0 \hat {\mathbf {j}}$
$\vec {\mathrm{V}} _ {\mathrm{RM}} = \vec {\mathrm{V}} _ {\mathrm{R}} - \vec {\mathrm{V}} _ {\mathrm{M}} = - 1 0 \hat {\mathrm{j}} - 5 \hat {\mathrm{i}}$
$\left| \vec {\mathrm{V}} _ {\mathrm{RM}} \right| = \sqrt {1 0 0 + 2 5} = 5 \sqrt {5} \mathrm{km/hr}$

$\vec{\mathrm{v}}_{\mathrm{RM}} = \vec{\mathrm{v}}_{\mathrm{R}} - \vec{\mathrm{v}}_{\mathrm{M}}$ $= -20\hat{\mathrm{j}} -30\hat{\mathrm{i}}$

Angle with vertical
$\tan \theta = \left(\frac {3 0}{2 0}\right)$
$\boxed {\theta = \tan^ {- 1} \left(\frac {3}{2}\right)}$

If man has horizontal velocity equal to horizontal component of rain velocity, then relative velocity in horizontal direction becomes zero, so rain will appear vertically.

$\vec{\mathrm{v}}_{\mathrm{rm}} = \vec{\mathrm{v}}_{\mathrm{r}} - \vec{\mathrm{v}}_{\mathrm{m}}$
$\vec {\mathbf {v}} _ {\mathrm{rm}} = - \mathbf {v} \hat {\mathbf {j}} - \mathbf {u} \hat {\mathbf {i}}$
$\tan \theta = \frac {\mathbf {u}}{\mathbf {v}}$

$\theta = \tan^ {- 1} \left(\frac {\mathbf {u}}{\mathbf {v}}\right)$

$v_{res} = \sqrt{v_{p}^{2} + v_{r}^{2}} = \sqrt{164}$ km/hr

$\sin \theta = \frac {5}{1 0} = \frac {1}{2}$
$\theta = 3 0 ^ {\circ}$
Angle with downstream = $90^{\circ} + 30^{\circ} = 120^{\circ}$

$t_{\mathrm{min}} = \frac{D}{v} = \frac{4}{2} = 2hr$

For shortest time he has to cross the river with swimming perpendicular to flow i.e. due north.

$t = \frac{d}{\sqrt{v^2 - v_R^2}} = \frac{4}{\sqrt{10^2 - 6^2}} = \frac{4}{8} = \frac{1}{2} hr$