Motion • Topic 2 of 3

Speed, Velocity & Acceleration

Knowing how far an object has moved is not enough — we usually want to know how fast it is moving. Speed tells us how quickly distance is covered. It is defined as the distance travelled per unit time: $\text{speed} = \dfrac{\text{distance}}{\text{time}}$. Speed is a scalar quantity (it has only magnitude). Its SI unit is the metre per second ($\text{m/s}$ or $\text{m s}^{-1}$); on Indian roads we usually use $\text{km/h}$, where $1\,\text{km/h} = \dfrac{5}{18}\,\text{m/s}$.

Because most motion is non-uniform, the speed keeps changing. So we use average speed, defined as the total distance divided by the total time taken: $v_{avg} = \dfrac{\text{total distance}}{\text{total time}}$. A bus from Jaipur to Delhi may stop at toll plazas and speed up on the highway, but its average speed over the trip is a single useful number.

Velocity is the speed of an object in a given direction — it is the displacement per unit time: $\text{velocity} = \dfrac{\text{displacement}}{\text{time}}$. Velocity is a vector quantity, so it has both magnitude and direction, and its SI unit is also $\text{m/s}$. Average velocity is the total displacement divided by total time. When velocity changes uniformly, the average velocity equals the mean of the initial and final velocities: $v_{avg} = \dfrac{u + v}{2}$.

When the velocity of an object changes, we say it is accelerating. Acceleration is the rate of change of velocity per unit time: $a = \dfrac{v - u}{t}$, where $u$ is the initial velocity and $v$ the final velocity. Acceleration is a vector and its SI unit is the metre per second squared ($\text{m/s}^2$). If the speed is increasing, acceleration is positive. If the speed is decreasing — like a car braking before a speed-breaker — the acceleration is negative; this negative acceleration is called retardation (or deceleration).

  • Speed $= \dfrac{\text{distance}}{\text{time}}$, scalar, unit $\text{m/s}$.
  • Velocity $= \dfrac{\text{displacement}}{\text{time}}$, vector, unit $\text{m/s}$.
  • Acceleration $a = \dfrac{v-u}{t}$, vector, unit $\text{m/s}^2$.
  • Retardation: negative acceleration, when an object slows down.
  • Conversion: $1\,\text{km/h} = \dfrac{5}{18}\,\text{m/s}$ and $1\,\text{m/s} = \dfrac{18}{5}\,\text{km/h}$.
Speed, Velocity and Acceleration relationshipsu = 5 m/sv = 25 m/sspeed = distance / timevelocity = displacement / timea = (v - u) / tVelocity grows from u to v, so a is positive
Solved Examples
1
Worked Example
A car travels $180\,\text{km}$ in $3\,\text{hours}$. Find its average speed in $\text{km/h}$ and in $\text{m/s}$.
Solution
  1. $v_{avg} = \dfrac{\text{distance}}{\text{time}} = \dfrac{180}{3} = 60\,\text{km/h}$.
  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
Convert $90\,\text{km/h}$ into $\text{m/s}$.
Solution
  1. Use $1\,\text{km/h} = \dfrac{5}{18}\,\text{m/s}$.
  2. $90 \times \dfrac{5}{18} = \dfrac{450}{18} = 25\,\text{m/s}$.

Answer: $90\,\text{km/h} = 25\,\text{m/s}$.

3
Worked Example
A car's velocity increases from $u = 10\,\text{m/s}$ to $v = 30\,\text{m/s}$ in $5\,\text{s}$. Find its acceleration.
Solution
  1. $a = \dfrac{v - u}{t} = \dfrac{30 - 10}{5}$.
  2. $= \dfrac{20}{5} = 4\,\text{m/s}^2$.
  3. It is positive, so the car speeds up.

Answer: $a = 4\,\text{m/s}^2$.

4
Worked Example
A bus moving at $20\,\text{m/s}$ is brought to rest in $4\,\text{s}$. Find its retardation.
Solution
  1. Here $u = 20\,\text{m/s}$, $v = 0$, $t = 4\,\text{s}$.
  2. $a = \dfrac{v - u}{t} = \dfrac{0 - 20}{4} = -5\,\text{m/s}^2$.
  3. The negative sign shows the bus is slowing down.

Answer: Retardation $= 5\,\text{m/s}^2$ (acceleration $= -5\,\text{m/s}^2$).

5
Worked Example
A man walks $400\,\text{m}$ in $5\,\text{minutes}$ and then $600\,\text{m}$ in $10\,\text{minutes}$. Find his average speed for the whole trip in $\text{m/s}$.
Solution
  1. Total distance $= 400 + 600 = 1000\,\text{m}$.
  2. Total time $= 5 + 10 = 15\,\text{min} = 15 \times 60 = 900\,\text{s}$.
  3. $v_{avg} = \dfrac{1000}{900} = 1.11\,\text{m/s}$.

Answer: Average speed $\approx 1.11\,\text{m/s}$.

6
Worked Example
An object moving in a straight line has $u = 4\,\text{m/s}$ and $v = 16\,\text{m/s}$ under uniform acceleration. Find its average velocity.
Solution
  1. For uniformly changing velocity, $v_{avg} = \dfrac{u + v}{2}$.
  2. $= \dfrac{4 + 16}{2} = \dfrac{20}{2} = 10\,\text{m/s}$.

Answer: Average velocity $= 10\,\text{m/s}$.

Key Points

  • Speed $= \dfrac{\text{distance}}{\text{time}}$ — scalar; unit $\text{m/s}$.
  • Velocity $= \dfrac{\text{displacement}}{\text{time}}$ — vector; unit $\text{m/s}$.
  • Acceleration $a = \dfrac{v-u}{t}$ — vector; unit $\text{m/s}^2$.
  • Negative acceleration (slowing down) is called retardation.
  • $1\,\text{km/h} = \dfrac{5}{18}\,\text{m/s}$ — a key conversion for numericals.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The SI unit of acceleration is:
Explanation: Acceleration is rate of change of velocity, so its unit is metre per second per second, i.e. m/s$^2$.
Q2.A train slows down from 20 m/s to 0 in 5 s. Its acceleration is:
Explanation: a = (v - u)/t = (0 - 20)/5 = -4 m/s$^2$. The negative sign indicates retardation.
Q3.72 km/h equals how many m/s?
Explanation: 72 x 5/18 = 360/18 = 20 m/s.
Q4.Which quantity is a vector?
Explanation: Velocity has both magnitude and direction, so it is a vector; speed and distance are scalars.
Practice more MCQs → 📝 Assignment · 30 marks