Motion in a Straight Line • Topic 2 of 3

Velocity & Acceleration

Knowing how far a particle has gone is only half the story; we also want to know how fast and how its motion is changing. Average speed is total path length divided by total time, a scalar: $\text{speed}_{avg} = \dfrac{\text{path length}}{\Delta t}$. Average velocity is total displacement divided by total time, a vector that carries the sign of $\Delta x$: $\bar{v} = \dfrac{\Delta x}{\Delta t} = \dfrac{x_2 - x_1}{t_2 - t_1}$. The SI unit of both is $\text{m/s}$, and the handy road conversion is $1\,\text{km/h} = \dfrac{5}{18}\,\text{m/s}$.

Average values hide the moment-to-moment detail. To capture the speedometer reading at one instant we shrink the time interval to zero. The instantaneous velocity is the limit of average velocity as $\Delta t \to 0$, which is exactly the calculus derivative of position with respect to time: $v = \lim_{\Delta t \to 0} \dfrac{\Delta x}{\Delta t} = \dfrac{dx}{dt}$. Geometrically, $v$ is the slope of the tangent to the position-time ($x$-$t$) graph at that instant. The magnitude of instantaneous velocity equals the instantaneous speed.

When velocity itself changes, the particle accelerates. Average acceleration is the change in velocity over the time taken: $\bar{a} = \dfrac{v_2 - v_1}{t_2 - t_1} = \dfrac{\Delta v}{\Delta t}$. Pushing $\Delta t \to 0$ gives the instantaneous acceleration, the derivative of velocity: $a = \dfrac{dv}{dt} = \dfrac{d^2x}{dt^2}$. Acceleration is a vector and its SI unit is $\text{m/s}^2$. On a velocity-time ($v$-$t$) graph, $a$ is the slope of the curve.

Signs deserve care in 1D. Acceleration is positive when velocity increases in the $+x$ sense and negative when velocity decreases — but a negative acceleration does not always mean slowing down. If a particle already moves in the $-x$ direction, a negative acceleration speeds it up. The reliable rule is: the particle speeds up when $v$ and $a$ have the same sign and slows down (retardation) when they have opposite signs.

A special, very common case is uniform acceleration, where $a$ is constant in both magnitude and direction. Free fall near the Earth's surface (ignoring air) is the classic example, with $a = g \approx 9.8\,\text{m/s}^2$ directed downward. Under uniform acceleration the velocity changes linearly with time, which is what makes the equations of motion in the next topic so simple.

  • Average velocity $\bar{v} = \dfrac{\Delta x}{\Delta t}$ (vector); average speed uses path length (scalar).
  • Instantaneous velocity $v = \dfrac{dx}{dt}$ = slope of the $x$-$t$ tangent.
  • Acceleration $a = \dfrac{dv}{dt} = \dfrac{d^2x}{dt^2}$ = slope of the $v$-$t$ graph, unit $\text{m/s}^2$.
  • Speeds up when $v$ and $a$ share a sign; slows down (retardation) when they differ.
  • Uniform acceleration: $a$ constant; velocity changes linearly with time.
x-t graph: slope of the tangent equals instantaneous velocityxt0x(t)tangentPt1slope of tangent at P = v = dx/dt
1
Worked Example
A car travels $240\,\text{km}$ in $4\,\text{hours}$ along a straight road. Find its average speed in $\text{km/h}$ and $\text{m/s}$.
Solution
  1. $\text{speed}_{avg} = \dfrac{240}{4} = 60\,\text{km/h}$.
  2. Convert: $60 \times \dfrac{5}{18} = \dfrac{300}{18} = 16.67\,\text{m/s}$.

Answer: $60\,\text{km/h} = 16.67\,\text{m/s}$.

2
Worked Example
A particle's position is $x = 4 + 6t - t^2$ (SI units). Find its velocity at $t = 2\,\text{s}$.
Solution
  1. $v = \dfrac{dx}{dt} = \dfrac{d}{dt}(4 + 6t - t^2)$.
  2. $v = 6 - 2t$.
  3. At $t = 2$: $v = 6 - 2(2) = 6 - 4 = 2\,\text{m/s}$.

Answer: $v = 2\,\text{m/s}$ at $t = 2\,\text{s}$.

3
Worked Example
For $x = 4 + 6t - t^2$, find the acceleration and the instant when the velocity becomes zero.
Solution
  1. $v = \dfrac{dx}{dt} = 6 - 2t$; $a = \dfrac{dv}{dt} = -2\,\text{m/s}^2$ (constant).
  2. Set $v = 0$: $6 - 2t = 0$.
  3. $t = 3\,\text{s}$.

Answer: $a = -2\,\text{m/s}^2$; velocity is zero at $t = 3\,\text{s}$.

4
Worked Example
A car accelerates from $u = 5\,\text{m/s}$ to $v = 23\,\text{m/s}$ in $6\,\text{s}$. Find its average acceleration.
Solution
  1. $\bar{a} = \dfrac{v - u}{t} = \dfrac{23 - 5}{6}$.
  2. $= \dfrac{18}{6} = 3\,\text{m/s}^2$.

Answer: $\bar{a} = 3\,\text{m/s}^2$.

5
Worked Example
A scooter moving at $20\,\text{m/s}$ slows uniformly to rest in $8\,\text{s}$. Find the acceleration and state whether it speeds up or slows down.
Solution
  1. $a = \dfrac{v - u}{t} = \dfrac{0 - 20}{8} = -2.5\,\text{m/s}^2$.
  2. Velocity is positive while acceleration is negative — opposite signs.
  3. Opposite signs mean the scooter slows down (retardation).

Answer: $a = -2.5\,\text{m/s}^2$; it slows down (retardation).

6
Worked Example
A man covers the first half of a straight trip at $30\,\text{km/h}$ and the second half (equal distance) at $60\,\text{km/h}$. Find his average speed for the whole trip.
Solution
  1. For equal distances, average speed $= \dfrac{2 v_1 v_2}{v_1 + v_2}$ (harmonic mean).
  2. $= \dfrac{2 \times 30 \times 60}{30 + 60} = \dfrac{3600}{90}$.
  3. $= 40\,\text{km/h}$.

Answer: Average speed $= 40\,\text{km/h}$.

Key Points

  • Average velocity $\bar{v} = \dfrac{\Delta x}{\Delta t}$ is a vector; average speed uses path length and is a scalar.
  • Instantaneous velocity $v = \dfrac{dx}{dt}$ equals the slope of the tangent to the $x$-$t$ graph.
  • Acceleration $a = \dfrac{dv}{dt} = \dfrac{d^2x}{dt^2}$ is the slope of the $v$-$t$ graph; unit $\text{m/s}^2$.
  • A particle speeds up when $v$ and $a$ have the same sign and slows down when they have opposite signs.
  • Uniform acceleration means $a$ is constant, so velocity changes linearly with time.
Tap an option to check your answer0 / 4
Q1.The instantaneous velocity is the slope of which graph?
Explanation: Since $v = \dfrac{dx}{dt}$, the slope of the tangent to the position-time graph gives instantaneous velocity.
Q2.For $x = 3t^2$ (SI units), the velocity at $t = 2\,\text{s}$ is:
Explanation: $v = \dfrac{dx}{dt} = 6t$, so at $t = 2$, $v = 12\,\text{m/s}$.
Q3.A body has positive velocity and negative acceleration. The body is:
Explanation: Opposite signs of $v$ and $a$ mean the speed is decreasing — retardation.
Q4.The SI unit of acceleration is:
Explanation: Acceleration is change of velocity per unit time, so its unit is $\text{m/s}^2$.