Description of the motion for a given interval of time.

u = velocity of the particle in the beginning of the interval.

v = velocity of the particle at the end of the interval.

a = constant acceleration of the particle.

s = displacement of the particle in the given time interval

x_{i} = co-ordinate of the particle along the line of motion, in the beginning of the interval.

x_{f} = coordinate of the particle along the line of motion at the end of the interval.

= the time interval.

The figure, describes the position, velocity and coordinate of a particle moving along x-axis at two different time and.(A) By definition:

a_{avg} = v-u/ t₂ – t₁

If acceleration of the particle is constant then

a_{avg} = a_{inst} = a

Hence, a = v – u/t, where, t = t₂ – t₁

(B) Displacement of the particle

s = x_{f} – x_{i} = <v> x t

If particle is moving with a constant acceleration then average velocity is defined as

<v> = (v + u)/2

s = (v + u)/2 x t

s = (u + at) + u/2 x t

As v = u + at from the equation

s = ut + ½ at²

(C) From equation a = (v – u)/t, t = (V – u)/a

If we put this value of t in equation s == [(V + u)/2] t we get

s = (v + u)/2 x (v – u)/a, v² = u² + 2as

**Kinematical equations: **The following derived equations are known as kinematical equations.

(i) v = u + at

(ii) s = ut + ½ at²

(iii) v² = u² + 2as.

**Note:**

(a) Any of the above equations can be used directly to solve the problems if and only if acceleration of the particle is a constant

(b) You must consider the sign of respective quantities.

**Example: **A particle is moving along a straight line with constant acceleration. At a point A on the line its velocity is equal to 4 m/s and at a point B its velocity becomes 6 m/s. If distance between these two points is 2 m find

(a) The acceleration of the particle

(b) The time taken by the particle to go to B from A.

(c) If at time t = 0, particle is at point A find the distance moved by the particle in the 5^{th} seconds.

**Solution:**

** **(a) In the given situation initial velocity u = 4 m/s and final velocity v = 6 m/s displacement s = 2m

∵ v² = u² + 2as

=> a = v² – u²/2s

=> (36m²/s² – 16 m²/s²)/2x2m = 5m/s²

(b) u = 4 m/s, v = 6 m/s and a = 5 m/s².

∵ v = u + at

=> t = v –u/a = (6m/s – 4m/s)/5m/s² = 0.4s

(c) Distance moved by the particle in 5 second is

=> S_{5} = 4m/s x 5s + ½ x 5m/s² x 25s² = 165/2 m = 82.5 m

Distance moved by the particle in 4 s is

=> S_{4} = 4m/s x 4s + ½ x 5m/s² x 16s² = 56m

Distance moved in the 5th second is S = S₅ – S₄

=> S = 82.5m – 56m = 26.5m