Acceleration

Acceleration is the rate at which velocity changes. It's a vector quantity, and in 1D it can be positive or negative.

Concept

Average acceleration over an interval: $\bar{a} = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{\Delta t}.$

Instantaneous acceleration: $a(t) = \frac{dv}{dt} = \frac{d^2 x}{dt^2}.$

A useful chain-rule form when acceleration depends on position: $a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\,\frac{dv}{dx}.$

This lets you write $a\,dx = v\,dv$, which integrates to give $v^2 - u^2 = 2as$ for uniform $a$.

Sign and Direction

• Acceleration along the direction of motion ($v$ and $a$ same sign): speed increases.
• Acceleration opposite to motion ($v$ and $a$ opposite signs): speed decreases ("deceleration").
• Acceleration zero ⇒ velocity constant ⇒ uniform motion.

Units

SI: $\text{m/s}^2$.

Worked Example

Q1: A particle's position is $x = 2t^3 - 6t^2 + 4t$ m. Find $v(t)$ and $a(t)$.

Solution: $v = \frac{dx}{dt} = 6t^2 - 12t + 4\,\text{m/s}$ $a = \frac{dv}{dt} = 12t - 12\,\text{m/s}^2$

At $t = 0$: $v = 4$ m/s, $a = -12$ m/s². The particle starts moving in the positive direction but decelerating.

Q2: Using $a = v\,dv/dx$: If $a = -3v$ (resistive deceleration) and $v_0 = 10$ m/s at $x = 0$, find $v$ at $x = 2$ m.

$v\,dv/dx = -3v \Rightarrow dv/dx = -3$. Integrate: $v = v_0 - 3x = 10 - 3(2) = 4$ m/s.

Common Confusions

• "Negative acceleration always means slowing down." — Only if velocity is positive. If velocity is also negative, a negative acceleration is speeding up in the negative direction.
• "Acceleration is zero when velocity is zero." — False. Throw a ball up; at the peak, $v = 0$ but $a = -g$ still.
• "Uniform velocity means uniform acceleration." — Uniform (constant) velocity means $a = 0$, not "uniform $a$."

Key Takeaways

• $a = dv/dt = d^2 x / dt^2$.
• Useful form: $a = v\,dv/dx$.
• Same sign as $v$ → speeding up; opposite → slowing down.
• $v = 0$ does not imply $a = 0$.
• For constant $a$, $v^2 - u^2 = 2as$.