Acceleration

Concept & Definition

Velocity tells us how position changes; acceleration goes one level deeper and tells us how velocity changes. Any time a body speeds up, slows down, or changes direction, it is accelerating — pressing you into your seat at a green light, forward against the belt while braking, or sideways through a sharp turn.

Like velocity, acceleration has both an average form over an interval and an instantaneous form at a single moment.

Units and Dimensions

The SI unit of acceleration is the meter per second squared ($\text{m/s}^2$), with dimensional formula $[\text{L}\,\text{T}^{-2}]$.

Mathematical Formulation

If a particle’s velocity changes by $\Delta v$ over the interval $\Delta t$, the average acceleration is:

Key Formula

Average Acceleration

$$\bar{a} = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{t_f - t_i}$$

where $\bar{a}$ is the average acceleration, $\Delta v$ is the change in velocity, and $\Delta t$ is the elapsed time.

Shrinking the interval to zero gives the instantaneous acceleration — the derivative of velocity, and therefore the second derivative of position:

Definition

Instantaneous Acceleration

$$a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt} = \frac{d^2x}{dt^2}$$

where $a$ is the instantaneous acceleration, $v(t)$ is the velocity, and $x(t)$ is the position of the particle as functions of time.

Reading the Signs

In 1D, acceleration and velocity with the same sign means the body is speeding up; opposite signs means it is slowing down. The sign of acceleration alone does not tell you which is happening — you must compare it with the sign of the velocity.

Common Pitfall

“Deceleration” in everyday speech means slowing down, but in physics there is only acceleration — a signed vector. A negative acceleration does not automatically mean slowing down: if the velocity is also negative, the body is actually speeding up in the negative direction.

Quick Check

A cyclist moving at 4 m/s reaches 12 m/s in 2 s. What is the average acceleration? If the velocity is v(t) = 4t, what is the instantaneous acceleration?

Reveal Explanation

Average: $\bar{a} = \dfrac{\Delta v}{\Delta t} = \dfrac{12 - 4}{2} = 4\ \text{m/s}^2$.

Instantaneous: differentiate the velocity, $a = \dfrac{dv}{dt} = \dfrac{d}{dt}(4t) = 4\ \text{m/s}^2$ — here the acceleration is constant, so average and instantaneous agree at every moment.