Straight line equal distance in equal interval of time.

Motion could be more complex

Car which starts from rest, starts to move, then it moves ot uniform motion,

How fast is position changing with time?

speed, include directional aspect $\to$ velocity

$\text{ Average velocity }=\frac{\text{ change in displacement }}{\text{ change in time }}$

If the displacement over time interval

$\mathrm{\Delta }t=\mathrm{\Delta }x$

$\text{ Then average velocity }$

$$$$$=\frac{\mathrm{\Delta }x}{\mathrm{\Delta }t}$

$$$\overline{V}=\frac{x_2-x_1}{t_2-t_1}$.

Units of average velocity $=\frac{\text{ Length }}{\text{ Time }}=\frac{L}{T}$

$SI=\frac{m}{s}$

Another unit $\frac{\mathrm{km}}{\mathrm{hr}}$

Average velocity $\to$ vector quantity motion along a straight line $\to$ direction is given by sign of displacement.

$\overline{v}=\frac{x_2-x_1}{t_2-t_1}$

$\mathrm{\Delta }t=t_2-t_1$

$PQ\to \text{ Straight line }$

$\text{ Slope of line }$

$\overline{v}$ can be positive, negative and zero.

${\text{ It }}_{\mathrm{\Delta }t\to 0}\frac{\mathrm{\Delta }x}{\mathrm{\Delta }t}=v$Instantaneous velocity rate of change of position with respect fo time.

$\frac{\mathrm{\Delta }x}{\mathrm{\Delta }t}\to$ approach tangent to the $x-t$ curve at $t_1$

$\text{ If }\mathrm{\Delta }t\to 0.$

tangent/slope of $x-t$ curve at $t=t$, gives velocity or instantaneous velocity at $t=t_1$

Motion could be more complex

Car which starts from rest, starts to move, then it moves at uniform motion,

Magnitude of Instantaneous Velocity = Instantaneous Speed.

Velocity may not be constant.

Rate of change of velocity with time as accelaration.

Motion could be more complex

Car which starts from rest, starts to move, then it moves at uniform motion,

Curve Upwards

$\leftarrow$ negative accelerant

straight line x - t

$\Rightarrow a=0$

$x\to t$

$\frac{d\overrightarrow{x}}{dt}=\overrightarrow{v}$

$\frac{dx}{dt}=v$

$\frac{dv}{dt}=a$

$x\to t$

$\frac{dx}{dt}=v$

$x-t$ curve slope = v

$\int dx=\int vdt$

$x_2-x_1$

Area under $v-t$ curve: displacement.

$\frac{dv}{dt}=a$

$\int dv=\int adt$

Slope of $v-t$ curve gives a Area under $a-t$ curve gives velocity $\to$ change of velocity.

If acceleration known as a funchon of $x$.

$\frac{dv}{dt}=a$

Chain Rule of Differentiation

$\frac{dv}{dt}=\frac{dv}{dx}\cdot \frac{dx}{dt}=\frac{vdv}{dx}$

$=\frac{1}{2}\frac{d}{dx}\left(v^2\right)$.

$\frac{1}{2}\frac{d}{dx}\left(v^2\right)=a$

$\int \frac{1}{2}d\left(v^2\right)=\int adx$

${\frac{1}{2}{v}^2|}_1^2=\int adx$

$=\frac{1}{2}(v_2^2-v_1^2)$

$=\int adx$.

$a$ is constant relation between displacement $x$, time taken $t$, initial velocity $v_0$, final velocity $v$ and acceleration $a$.

acceleration = $a\leftarrow$ constant

$a=\frac{v-v_0}{t}$

$v=v_0+at$

Area of triangle = $\frac{1}{2}(v-v_0)t$

displacement

Area under $v-t$ curve

Area $=v_{0}t+\frac{1}{2}(v-v_0)t$

displacement = $v_{0}t+\frac{1}{2}a{t}^2$

$a$ is constant relation between displecemend $x$, time taken $t$, initial velocity $v_0$, firal velocity $v$ and acceleration $a$.

acceleration = $a\leftarrow$ constant

$a=\frac{v-v_0}{t}$

$v=v_0+at$

$\frac{1}{2}(v-v_0)t$

displacement

Area under $v-t$ curve.

Area $=v_{0}t+\frac{1}{2}(v-v_0)t$.

$x=v_{0}t+\frac{1}{2}a{t}^2$

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

$x=\frac{(v+v_0)}{2}\frac{(v-v_0)}{a}$

$x=\frac{v^2-v_0^2}{2a}$

$v^2=v_0^2+2ax$.

$a=\frac{v-v_0}{t}$

$v=v_0+at$

$\frac{1}{2}(v-v_0)t$

displacement

Area under $v-t$ curve

Area $=v_{0}t+\frac{1}{2}(v-v_0)t$

displacement

$=v_{0}t+\frac{1}{2}a{t}^2$.

$x=v_{0}t+\frac{1}{2}a{t}^2$

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

$x=\frac{(v+v_0)}{2}\frac{(v-v_0)}{a}$

$x=\frac{v^2-v_0^2}{2a}$

$v^2=v_0^2+2ax$.

$v=v_0+at$ $⟶(x-x_0)$ missing

$x=v_{0}t+\frac{1}{2}a{t}^2$ $⟶v$ missing

$v^2=v_b^2+2ax$ $⟶t$ missing

Valid only if $a=$ constant

$x\to$ displacement

$x_0\ne 0$

In above formulae $x_0=0$.

If $x_0\ne 0\to x$ replaced by $(x-x_0)$.

Sign of $a$

If velocity increasing in direction of positive $x,$

$a$ is positive.

If velocity decreasing in direction of positive $x$,

$a$ is negative (retardation).

Body falling under influence of gravity near earth’s surface acceleration is constant $=g$ from the body toward the surface of earth.

$g=9.8\frac{m}{s^2}$

or

$g=10\frac{m}{s^2}$

-ve sign downward directon.

+ve sign upward diredton.

$\downarrow y$ If you choose y as downwards

$a=+g$