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Formulas/physics/Kinematics/Uniform Motion

Uniform Motion

Position of a body moving at constant velocity.
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Derivation

What this formula says

A body is at position x0x_0 at time t=0t = 0. It moves with a constant velocity vv. After time tt, its position is:

x=x0+vtx = x_0 + vt

Its new position is its starting position plus how far it has travelled. That is all.

What uniform motion means

Uniform motion means the velocity is constant — same magnitude, same direction, at every instant. This requires zero net force and no change in direction.

A car cruising on a straight highway at a steady speed is approximately uniform motion. A planet in a circular orbit moves at constant speed but is not uniform motion — its direction changes continuously.

Derivation

Velocity is defined as the rate of change of position:

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

Since vv is constant, separate variables:

dx=vdtdx = v \, dt

Integrate both sides — position from x0x_0 to xx, time from 00 to tt:

x0xdx=0tvdt\int_{x_0}^{x} dx = \int_0^t v \, dt

Left side: xx0x - x_0

Right side (vv constant, comes outside): vtvt

xx0=vtx - x_0 = vt

x=x0+vt\boxed{x = x_0 + vt}

Without calculus

If you travel at constant velocity vv for time tt, distance covered is vtvt. Add this to starting position:

final position=initial position+distance covered=x0+vt\text{final position} = \text{initial position} + \text{distance covered} = x_0 + vt

Sign convention

QuantityPositiveNegative
x0x_0Started right of originStarted left of origin
vvMoving rightMoving left
xxFinal position right of originFinal position left of origin

Example: Body starts at x0=5x_0 = -5 m, moves at v=+3v = +3 m/s for t=4t = 4 s:

x=5+3×4=+7 mx = -5 + 3 \times 4 = +7 \text{ m}

Started 5 m left of origin, ended 7 m right of it.

Note
This formula applies only when velocity is truly constant. The moment any acceleration is present, use the equations of motion instead.