Rectilinear Motion of Particles [Kinematics]

Displacement (s) = $x_t – x_o$


$d↙{(0 -> t)} = |x_y – x_o|$

Instantaneous velocity = dx/dt

Displacement ≦ Distance
Velocity ≦ speed

a = dv/dt OR a = a ${dv}/{ds}$

Jerk: Rate of change of acceleration w.r.t time

$J = {da}/{dt}$

Types of motion and kinematic equations

(1) Uniform motion: v is constant and a = 0
s= v x t

(2)Uniformly accelerated motion: a os constant and J = 0
V = u + at, s = ut + $1/2$ a $t^2$
$v^2 = u^2 + 2as$

(3)Motion under gravity: g is constant

v = u – gt , $s = ut – 1/2 gt^2$
$v^2 = u^2 – 2 gs$

Distance travelled in $n^{th}$ second

For U.A.M: $s^{nth} = u + a/2 (2n-1) $

For M.U.G: $s^{nth} = u – g/2 (2n -1)$

For M.U.G
$v = √{2gh}$
$t = √{{2h}/{g}}$
Max. Height wen particle projected from ground, $h = {v^2}/{2g}$

h α (square of velocity)

Motion diagram

v – t diagram:

(a) Slope of v-t diagram = ${dv}/{dt}$ = acceleration
(b) Area under v – t digram = change in position ($x_2 – x_1$)
(c) For velocity to be maximum ${dv}/{dt}$ = a = 0

a – t digram:
(a) Slope of a-t diagram = ${da}/{dt}$ = jerk

(b) Area under a-t diagram = change in velocity ($v_2 – v_1$)

(c) To find position, use moment equation
$x_t = x_0 + v_o t + M_t$

x – t digram:
Slope of x – t diagram = ${dx}/{dt}$ = velocity

For x to be maximum ${dx}/{dt}$ = v = 0

v – x diagram:
Property: Acceleration = velocity x slope of v.x diagram

a – x diagram


Area under a – x diagram = ${{v^2_2} – {v^2_1}}/{2}$

Note: when $v^2$ α s, acceleration of particle is uniform