$\text { Relative Velocity and examples }$

Uniform Acceleration 'a'

$v=v_0+a t$

$x-x_0=v_0 t+\frac{1}{2} a t^2$

$v^2=v_0^2+2 a\left(x-x_0\right)$

$x-x_0=\frac{1}{2}\left(v_0+v\right) t$

$x-x_0=v t-\frac{1}{2} a t^2$

$v, v_0, a, t, x-x_0$

$\downarrow x$

$a = + 9$

$= 10\text {m}/\text {s}^2$

$+ = 0.2$

$\left(x-x_0\right)=V_0 t+\frac{1}{2} a t^2$

Use this formula from $1$ to $2$

$1 m=V_1(0.2)+\frac{1}{2} g \times(0.2)^2$

$\hspace {1cm} \uparrow \hspace {1.4cm} \uparrow 10$

$1 =0.2 V_1+5(0.2)^2$

$\Rightarrow v_1=4 \mathrm{~ m} / \mathrm{s}$

$\text { Total time from } 1 \text { to } 3=1+0.2 = 1.2\text {s}$

$S_1 =4(1.2)+\frac{1}{2} \times 10 \times(1.2)^2 =4.8+7.2 =12 \mathrm{~ m}$

$S_0$ = ? (x from top of the building to top of the window)

$\text { Know } V, V_0, a, \left(x-x_0\right)$

$V^2=v_0^2+2 a ~ S_0$

$16=0+2 \times 10 \times S_0$

$\Rightarrow S_0=\frac{16}{20}=0.8 \mathrm{~ m}$

$\text { Total height }=12+0.8=12.8 \mathrm{~ m}$

$\downarrow x$

$a = + 9$

$= 10\text {m}/\text {s}^2$

$+ = 0.2$

$\left(x-x_0\right)=V_0 t+\frac{1}{2} a t^2$

use this formula from $1$ to $2$

$1 m=V_1(0.2)+\frac{1}{2} g \times(0.2)^2$

$\hspace {1cm} \uparrow \hspace {1.4cm} \uparrow 10$

$\left.\begin{array}{l} A C=S_1 \\ C B=S_2 \end{array}\right\} \quad S_1+S_2=l$

$\text { Let velocity at } C \text { be } V_C$

$V_C^2=V_A^2+2 a_1 s_1 \rightarrow S_1 = \frac{V_C^2}{2a_1}$

$V_B^2=V_C^2-2 a_2 s_2\rightarrow S_2 = \frac{V_C^2}{2a_2}$

$l=\frac{V_c^2}{2 a_1}+\frac{V_c^2}{2 a_2}$

$V_c^2\left(\frac{1}{2 a_1}+\frac{1}{2 a_2}\right)=l$

$V_c^2=\frac{l\left(2 a_1 a_2\right)}{\left(a_1+a_2\right)}$

$\text { Time } = ?$

$V_c = V_A^0 + a_1 t_1$

$t_1 \rightarrow\text { Time from } A \text { to } C$

$V_B^0=V_C-a_2 t_2$

$t_1 = \frac{V_c}{a_1}, \quad t_2 = \frac{ V_c}{a_2}$

$\text { Total time } = \left(t_1 + t_2\right) = V_c \left(\frac{1}{a_1}+\frac{1}{a_2}\right)$

$V_c = \sqrt{\frac{ l\left(2a_1 a_2\right)}{\left(a_1+a_2\right)}}$

$t = \sqrt {\frac{2l\left(a_1+a_2\right)}{a_1 a_2}}$

$\text { Total distance = l }$

$\text { find } \left (t_1 + t_2\right)$

$t_1=\frac{v_c}{a_1}$

$t_2=\frac{v_c}{a_2}$

$\left(t_1+t_2\right)=\frac{v_c}{a_1}+\frac{v_c}{a_2}=v_c\left(\frac{1}{a_1}+\frac{1}{a_2}\right)$

$v_c=\frac{l}{2 t} \Rightarrow t_2$

$t=\frac{l}{2 t}\left(\frac{1}{a_1}+\frac{1}{a_2}\right)$

$\Rightarrow t^2 = \frac{l}{2} \left(\frac{1}{a_1}+\frac{1}{a_2}\right)$

$\text { Find } t_1 \text { to } t_2:$

$\left(t_1+t_2\right)=\frac{v_c}{a_1}+\frac{v_c}{a_2}=v_c\left(\frac{1}{a_1}+\frac{1}{a_2}\right)$

$v_c=\frac{l}{2 t} \Rightarrow t_2$

$t=\frac{l}{2 t}\left(\frac{1}{a_1}+\frac{1}{a_2}\right)$

$\Rightarrow t^2 = \frac{l}{2} \left(\frac{1}{a_1}+\frac{1}{a_2}\right)$

Position of a point depends on the reference frame from where it is being measured.

Position is a frame dependent quantity.

$\rightarrow$ Velocity and Acceleration also become frame dependent quantities.

Let us consider a point $P$ moving in a straight line.

An observer attached to frame $A$

$x_{\text {PA}}$ location of $P$ as observed by observer attached to frame $A$.

$x_{P A}=x_{P B}+x_{B A}$

$\frac{d x_{P A}}{d t}=\frac{d x_{P B}}{d t}+\frac{d x_{B A}}{d t}$

$v_{P A}=\hspace {0.1cm}v_{P B}\hspace {0.1cm}+v_{B A}$

$v_{P A} \rightarrow \text { Velocity of } P \text { as observed from frame } A$

$v_{P B} \rightarrow\text { Velocity of } P \text { as observed from frame } B$

$v_{B A} \rightarrow\text { Velocity of } B \text { as observed from frame }A$

$A \rightarrow \text{ Fixed to ground }$

$A \text { may be dropped. }$

$v_P = v_{PB} + v_B$

$v_{PB} = v_P - v_B$

$v_{PB} \rightarrow \text { Velocity of }P \text { as observed by frame } B$

$v_P \rightarrow \text { Velocity of } P$

$v_B \rightarrow \text { Velocity of } B$

$\text { If both frame }A \text { and }B \text { move with constant velocities,}$

$a_A = a_B = 0$

$a_{PA} = a_{PB} + a_{BA}$

$a_{BA} \rightarrow \left(a_B - a_A\right)$

$a_{PA} = a_{PB}$

$v_{P A}=v_{P B}+v_{B A}$

Caution: Relation valid for low speeds,

$v_{P B}$, $v_{B A} \ll C$

$C= \text { Speed of light. }$

Time in which the person will move from bottom to top of escalator if he moves up both his speed $C$ ecalator is also moving.

Person $P: \hspace {0.5cm} v_p=\frac{l}{t_1}$

Escalator moves $\rightarrow v_P =$ Velocity of person

w.r.t. Escalator

$\text { Person moves up stopped } \rightarrow t_1 \text { escalator, }$

$\text { Bottom to top } \rightarrow t_2$

$\text { Person moves with his speed on moving escalator} \rightarrow \text { find } t_3$.

$\text { Length of the escalator } =l$

$v_p=\frac{l}{t_1}$

When Person moves on the escalator

$\left.v_P \rightarrow v_P\right|_e$

On a moving escalator with person moving

$v_P|_e=\frac{l}{t_1}$

$v_{\text {escalator }}=\frac{l}{t_2}$

Let $v_c$ be speed of cyclist

$t_1 \rightarrow$ when buses pass in direction of cyclist

$t_2 \rightarrow$ when buses pass in rerverse direction

$v_B \rightarrow$ speed of bus.

Method - 1:

Distance travelled by bus $2$ when it crosses cyclist

$\left(\text {from } t = 0\right)$

$(1) \rightarrow \left ( v_BT + v_c ~ t_1 \right) = v_B t_1$

$(2) v_BT = v_c t_2 + V_B t_2$

$(1) v_BT = \left(v_B - v_c\right) t_1$

$(2) v_BT = \left(v_B + v_c\right) t_2$

$(1) = (2) \rightarrow v_B.$

Method-2

Observe motion of two bus in the frame of cyclist.

$\text { When bus 2 cyclist travel in the same direction }$

$v_B \longrightarrow v_c\\ \longrightarrow \\ B$

$\text { Velocity of bus as seen by cyclist. }$

$v_{BC} = v_B - v_c$

$\text { To the cyclist: the bus travles. } \left(v_BT\right)$

$\text { Time taken }= t_1$

$\left(v_{B C}\right) t_1$ = Distance travels by bus as-seen in frame of cyclist = $v_B T$

$(v_B-v_c) t_1= v_BT$

Velocity of bus in ground frame =- $v_B$

Velocity of bus as seen cyclist } = $v_{B / G}-v_{c / G} = -v_B-v_c$

$v_{B / c}=-\left(v_B+v_c\right)$

$\text { In reference frame of cyclist }$

$\text { Bus travels } - \underbrace{v_BT} \text { before it or crosses him. }$

$\left(v_{B /C}\right) t_2 =-v_B T$

$-\left(v_B+v_C\right) t_2 =-v_B T \quad (2)$