When we study kinematics, we do not go in the details of what is causing the motion, but we just analyze the motion.

Planar motion.

To describe position of a point, coordinate system.

Cartesian axes (x, y)

We take two mutually perpendicular directions.

x and y are at an angle of 90 degrees to each other.

The intersection of these is called the origin represented by O.

Any point P, which is at a location on the cartesian plane, is described by the coordinates of x and y.

Position $P$ of a point is given by coordinates$(x, y)$

$\overrightarrow{O P}$ from origin to length from origin to position of P.

$\overrightarrow{O P} \rightarrow$ Position vector

Lenght $OP \rightarrow$ Magnitude.

Vector has 2 characteristics

1) Magnitude (length)

2) Direction

Concept of a derivative

$y=f(x)$, $y$ is a function of $x$.

For different values of x, we have different values of y.

$P(x, y)$, $Q(x+\Delta x, y+\Delta y)$

Straight line joining $P Q$

$\tan \theta=\frac{\Delta y}{\Delta x}$.

Q approach $P \rightarrow \Delta x \rightarrow 0 , \Delta y \rightarrow 0$. But $\frac{\Delta y}{\Delta x}$ will not approach zero.

PQ approach tangent to the and at point $P$.

If denote slope of line by $m$.

$m=\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}$

Define derivative of a function $y$ where $x$,

Where $y=f(x)$

$\frac{d y}{d x}=\frac{d f(x)}{d x}=\underset{\Delta x \rightarrow 0}{\operatorname{lim}} \frac{f(x+\Delta x)-f(x)}{\Delta x}$

2 functions, $u(x), v(x)$

$\frac{d(u+v)}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$

$\frac{d(u v)}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}$

$\frac{d(\frac{u}{v})}{d x}=\frac{1}{v^2} \frac{d u}{d x}-u \frac{d v}{d x}$

$\frac{d x^n}{d x}=n x^{n-1}$

If u = u(x),

Chain Rule: $\frac{d u^n}{d x}=n u^{n-1} \frac{d u}{d x}$

Each interval has a length $\Delta {x}_{i}$

$\Delta A_i=f\left(x_i\right) \Delta x$

$\text { Total Area }=\sum \Delta A_i$

Total Area $=\sum \Delta A_ i=\sum _ {i=1}^N f\left(x_i\right) \Delta x$

$N \rightarrow \infty$, $\sum \rightarrow \text { integral } \int$

$A=\int_{x=a}^ {x=b} f(x) d x =\int _ {a}^{b} f(x) d x$.

Integration $\rightarrow$ Inverse of differentiation.

$g(x)$ such that $\frac{d g(x)}{d x}=f(x)$

$g(x)=\int f(x) d x$

$\int_a^b f(x) d x=g(b)-g(a)$

$\int x^n d x=\frac{x^{n+1}}{n+1}+c$

$\int \frac{1}{x} d x=\ln x+c$

$\int \sin x d x=-\cos x+c$

$\int \cos x d x=\sin x+c$

Particle is on entity of very small size (point) but of a finite mass.

Treat cricket ball treated as a particle.

If we need overall estimates of path moved by body (without (bothering about details of different parts of the body) $\rightarrow$ lie can treat body as a particle.

Distance moved by body $\gg$ size.

In a reference frame:

Device to measure length device to measure time (clock) observe motion of a point $P$ in this reference frame.

Reference frame 1:

Ground car is moving.

Reference frame 2:

Car someone sitting on seat of the car.

Is any reference from which is absolutely stationary?

Reference frame fixed to car $\rightarrow$ moving w.r.t.ground.

Reference frame fixed to ground, $\rightarrow$ rotating with the earth.

Reference frame to centre of sun.

Path At time t

$\overrightarrow{O P}=\vec{r}(t)$

$t+\Delta t$

Particle at P'

$\overrightarrow{O P^{\prime}}=\overrightarrow{r^{\prime}}(t+\Delta t)$

At time $t$, point $P$.

$\overrightarrow{PP'} \rightarrow$ Displacement vector.

$\operatorname{lim}_{\Delta t \rightarrow 0} \frac{\vec{r}(t+\Delta t)-\vec{r}(t)}{\Delta t}$

$=\vec{v}(t)$

$\vec{V}_p(\text { at time t})$ = Derivative of position vector

Acceleration of point $P$ :

$\overrightarrow{a_p}= \operatorname{lim}_{\Delta t \rightarrow 0} \frac{{\overrightarrow{V_p}(t+\Delta t)-\overrightarrow{V_p}(t)}}{\Delta t}$.

Particle moves along a straight line.

Align ${x}-$ axis along the straight line.

Position:

Location in space where the particle is at time t.

t= 0s Particle at O

t= 1s Particle at P

t= 2s particle at Q

t= 3s Particle at P

Displacement $\equiv \frac{\Delta x}{\Delta t}=\frac{x_2-x_1}{t_2-t_1}$

$x_ 2>x_1 \rightarrow$ positive axis

$x_ 2< x _ 1 \rightarrow$ negative axis

Displacement can be positive and negative.

Positive $\rightarrow$ if particle moves along ${x}$-axis.

Negative if paricle move against x axis.

Moving along $-x$ axis.

Displacement is a vector quantity.

In one dimensional motion, direction of displacement is given by the sign, is positive then it is along + x axis and if it is negative then it is along - x axis.