# Three Equations of Motion

**Three Equations of Motion:**

1. The Equation for Velocity – Time Relation

v = u + at

s = ut + 1/2 at^2

3.The Equation for the Position – Velocity Relation

2a s = v^2 – u^2

Where u: initial velocity

a: uniform acceleration

t: time

v: final velocity

s: distance travelled in time t

**Deriving the Equations of Motion Graphically**

Study the graph above. The line segment PN shows the relation between velocity and time. Initial velocity, u can be derived from velocity at point P or by the line segment OP

Final velocity, v can be derived from velocity at point N or by the line segment NR

Also, NQ = NR – PO = v – u

Time interval, t is represented by OR, where OR = PQ = MN

**Deriving the Equation for Velocity** – Time Relation

Acceleration = Change in velocity / time taken

Acceleration = (final velocity – initial velocity) / time

a = (v – u)/t

so, at = v – u

**v = u + at**

**2. Deriving Equation for Position**– Time Relation

We know that, distance travelled by an object = Area under the graph

So, Distance travelled = Area of OPNR = Area of rectangle OPQR + Area of triangle PQN

s = (OP * OR) + (PQ * QN) / 2

s = (u * t) + (t * (v – u) / 2)

**s = ut + 1/2 at2 ** [because at = v – u]

**3. Deriving the Equation for Position **– Velocity Relation

We know that, distance travelled by an object = area under the graph

So, s = Area of OPNR = (Sum of parallel sides * height) / 2

s = ((PO + NR)* PQ)/ 2 = ( (v+u) * t)/ 2

2s / (v+u) = t [equation 1]

Also, we know that, (v – u)/ a = t [equation 2]

On equating equations 1 and 2, we get,

2s / (v + u) = (v – u)/ a

2as = (v + u) (v – u)

2 a s = v^2 – u^2