Velocity, speed, and acceleration

If a particle’s position at time $t$ is given by the vector-valued function $\mathbf{r}(t)$, then

• Velocity is $\mathbf{v}(t)={\mathbf{r}}^{\prime }(t)$, a vector: direction of motion times speed.
• Speed is $|\mathbf{v}(t)|=|{\mathbf{r}}^{\prime }(t)|$, a scalar: how fast.
• Acceleration is $\mathbf{a}(t)={\mathbf{v}}^{\prime }(t)={\mathbf{r}}^{\prime \prime }(t)$, a vector: rate of change of velocity.

Difference between speed and velocity

In everyday speech the two words get swapped, but here they’re sharply different: velocity is a vector (with a direction), speed is its magnitude (a non-negative scalar). A particle moving in a circle at constant speed has velocity that keeps changing direction, so it’s accelerating even though speed is constant.

Worked example: the helix

$\mathbf{r}(t)=⟨\cos t,\sin t,t⟩$.

$\mathbf{v}(t)=⟨-\sin t,\cos t,1⟩$.

Speed: $|\mathbf{v}|=\sqrt{{\sin }^{2}t+{\cos }^{2}t+1}=\sqrt{2}$. Constant.

$\mathbf{a}(t)=⟨-\cos t,-\sin t,0⟩$. Magnitude $|\mathbf{a}|=1$.

The acceleration points horizontally inward, toward the $z$-axis: the centripetal acceleration of circular motion in the $xy$-plane, happening at the same time as the steady $\hat{\mathbf{z}}$ drift. Constant speed, nonzero acceleration. The helix shows exactly why those two have to be distinguished.

Displacement vs. distance traveled

Two different integrals.

Net displacement from time $a$ to $b$:

${\int }_a^b\mathbf{v}(t)dt=\mathbf{r}(b)-\mathbf{r}(a),$

a vector. This is the straight-line vector from start to end, ignoring the path.

Distance traveled (path length) from $a$ to $b$:

${\int }_a^b|\mathbf{v}(t)|dt,$

a non-negative scalar. This is the Arc length of the path.

For a particle that returns to its starting point, the net displacement is $0$ but the distance traveled is the total path length, usually nonzero.

Decomposition of acceleration

Acceleration splits into tangential and normal components:

$\mathbf{a}=a_T\mathbf{T}+a_N\mathbf{N},$

where $\mathbf{T}$ is the unit tangent, $\mathbf{N}$ is the principal normal, $a_T=\frac{d}{dt}|\mathbf{v}|$ (rate of change of speed), and $a_N=|\mathbf{v}{|}^2/R$ with $R$ the radius of curvature (centripetal).

For uniform circular motion, $a_T=0$ (constant speed) and $a_N=v^2/R$, pure centripetal. For straight-line motion with changing speed, $a_N=0$ and $a_T=dv/dt$, pure tangential.