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Distance, displacement and velocity

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Introduction to Motion: Distance, Displacement, and Velocity

Motion is a fundamental concept in physics that describes how objects change their position over time. To understand motion clearly, it is essential to distinguish between distance, displacement, and velocity. These terms help us describe how far an object has moved, in which direction, and how fast.

Before diving into these concepts, let's understand two important types of quantities used in physics:

  • Scalar quantities: These have only magnitude (size or amount) but no direction. Examples include distance, speed, and time.
  • Vector quantities: These have both magnitude and direction. Examples include displacement, velocity, and acceleration.

Recognizing whether a quantity is scalar or vector is crucial because it affects how we calculate and interpret motion.

Distance and Displacement

Imagine you walk from your home to a nearby park. You might take a winding path through streets and alleys. The total length of this path you walked is your distance. It tells you how much ground you covered, regardless of direction.

Now, if you were to fly like a bird straight from your home to the park, the shortest straight-line distance between these two points is called displacement. Displacement considers direction and is represented as an arrow pointing from the starting point to the ending point.

Start End Displacement Distance (curved path)

Key points:

  • Distance is the total length of the path traveled. It is always positive or zero and does not depend on direction.
  • Displacement is the straight-line vector from the initial to the final position. It can be zero if the start and end points are the same, even if distance traveled is not zero.

Speed and Velocity

When we talk about how fast something is moving, we use speed and velocity. Although they sound similar, they have important differences.

Speed is the rate at which distance is covered. Since distance is scalar, speed is also a scalar quantity. For example, if a car covers 100 kilometers in 2 hours, its average speed is 50 km/h.

Velocity is the rate at which displacement changes with time. Because displacement is a vector, velocity is also a vector quantity and includes direction. For example, a car moving 50 km/h east has a velocity of 50 km/h east.

Speed = distance/time (scalar) Velocity = displacement/time (vector)

Velocity can be further classified as:

  • Average velocity: Total displacement divided by total time.
  • Instantaneous velocity: Velocity at a specific instant of time.

Understanding the difference between speed and velocity is important because an object can have a high speed but zero velocity if it returns to its starting point (displacement zero).

Distance

\[d = \sum \text{path segments}\]

Total length of the path traveled, scalar quantity.

d = distance (m)

Displacement

\[\vec{s} = \vec{r}_{final} - \vec{r}_{initial}\]

Vector from initial to final position.

\(\vec{s}\) = displacement (m)
\(\vec{r}\) = position vector (m)

Speed

\[v = \frac{d}{t}\]

Rate of change of distance with time, scalar.

v = speed (m/s)
d = distance (m)
t = time (s)

Velocity

\[\vec{v} = \frac{\vec{s}}{t}\]

Rate of change of displacement with time, vector.

\(\vec{v}\) = velocity (m/s)
\(\vec{s}\) = displacement (m)
t = time (s)

Average Velocity

\[\vec{v}_{avg} = \frac{\vec{s}_{total}}{t_{total}}\]

Total displacement divided by total time.

\(\vec{v}_{avg}\) = average velocity (m/s)
\(\vec{s}_{total}\) = total displacement (m)
\(t_{total}\) = total time (s)

Worked Examples

Example 1: Calculating Displacement and Distance Easy
A particle moves 3 km east and then 4 km north. Calculate the total distance traveled and the displacement.

Step 1: Calculate total distance.

Distance is the sum of all path segments:

\( d = 3\, \text{km} + 4\, \text{km} = 7\, \text{km} \)

Step 2: Calculate displacement.

Displacement is the straight-line distance from start to end. Since the motion is east then north, displacement forms the hypotenuse of a right triangle.

Using Pythagoras theorem:

\( s = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\, \text{km} \)

Step 3: Direction of displacement is northeast, at an angle \( \theta \) from east:

\( \theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ \) north of east

Answer: Distance traveled = 7 km; Displacement = 5 km at 53.13° north of east.

Example 2: Average Velocity from Displacement and Time Easy
A runner covers 1000 meters north in 5 minutes. Find the average velocity.

Step 1: Convert time to seconds.

\( 5\, \text{minutes} = 5 \times 60 = 300\, \text{seconds} \)

Step 2: Use average velocity formula:

\( \vec{v}_{avg} = \frac{\vec{s}}{t} = \frac{1000\, \text{m north}}{300\, \text{s}} = 3.33\, \text{m/s north} \)

Answer: Average velocity is 3.33 m/s north.

Example 3: Velocity with Direction Change Medium
A car moves 10 km east and then 6 km west in 2 hours. Find the average velocity.

Step 1: Calculate total displacement.

Displacement = final position - initial position.

East is positive, west is negative, so:

\( s = 10\, \text{km} - 6\, \text{km} = 4\, \text{km east} \)

Step 2: Calculate average velocity.

\( \vec{v}_{avg} = \frac{4\, \text{km east}}{2\, \text{hours}} = 2\, \text{km/h east} \)

Step 3: Calculate average speed.

Total distance = 10 km + 6 km = 16 km

Average speed = \( \frac{16\, \text{km}}{2\, \text{hours}} = 8\, \text{km/h} \)

Answer: Average velocity = 2 km/h east; Average speed = 8 km/h.

Example 4: Relative Velocity in One Dimension Medium
Two trains are moving towards each other on parallel tracks. Train A moves at 60 km/h and Train B at 40 km/h. Find the relative velocity of Train A with respect to Train B.

Step 1: Define directions.

Assume Train A moves east at +60 km/h, Train B moves west at -40 km/h.

Step 2: Relative velocity formula:

\( \vec{v}_{A/B} = \vec{v}_A - \vec{v}_B \)

\( = 60\, \text{km/h} - (-40\, \text{km/h}) = 60 + 40 = 100\, \text{km/h} \)

Answer: The relative velocity of Train A with respect to Train B is 100 km/h.

Example 5: Velocity-Time Graph Interpretation Hard
A velocity-time graph shows a car accelerating uniformly from 0 m/s to 20 m/s in 5 seconds, then moving at constant velocity for 10 seconds, and finally decelerating uniformly to rest in 5 seconds. Find the total displacement and average velocity during the 20 seconds.

Step 1: Break the motion into three parts:

  • Acceleration: 0 to 20 m/s in 5 s
  • Constant velocity: 20 m/s for 10 s
  • Deceleration: 20 m/s to 0 in 5 s

Step 2: Calculate displacement for each part using area under velocity-time graph.

Part 1 (triangle):

\( \text{Displacement}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 20 = 50\, \text{m} \)

Part 2 (rectangle):

\( \text{Displacement}_2 = \text{velocity} \times \text{time} = 20 \times 10 = 200\, \text{m} \)

Part 3 (triangle):

\( \text{Displacement}_3 = \frac{1}{2} \times 5 \times 20 = 50\, \text{m} \)

Step 3: Total displacement:

\( s_{total} = 50 + 200 + 50 = 300\, \text{m} \)

Step 4: Average velocity:

Total time = 5 + 10 + 5 = 20 s

\( v_{avg} = \frac{s_{total}}{t_{total}} = \frac{300}{20} = 15\, \text{m/s} \)

Answer: Total displacement = 300 m; Average velocity = 15 m/s.

Formula Bank

Distance
\[ d = \sum \text{path segments} \]
where: \( d \) = distance (m)
Displacement
\[ \vec{s} = \vec{r}_{final} - \vec{r}_{initial} \]
where: \( \vec{s} \) = displacement (m), \( \vec{r} \) = position vector (m)
Speed
\[ v = \frac{d}{t} \]
where: \( v \) = speed (m/s), \( d \) = distance (m), \( t \) = time (s)
Velocity
\[ \vec{v} = \frac{\vec{s}}{t} \]
where: \( \vec{v} \) = velocity (m/s), \( \vec{s} \) = displacement (m), \( t \) = time (s)
Average Velocity
\[ \vec{v}_{avg} = \frac{\vec{s}_{total}}{t_{total}} \]
where: \( \vec{v}_{avg} \) = average velocity (m/s), \( \vec{s}_{total} \) = total displacement (m), \( t_{total} \) = total time (s)

Tips & Tricks

Tip: Always distinguish between scalar and vector quantities by checking if direction matters.

When to use: When solving problems involving distance vs displacement or speed vs velocity.

Tip: Use Pythagoras theorem to find displacement when motion is along perpendicular directions.

When to use: When an object moves in two perpendicular directions, e.g., east then north.

Tip: Convert all units to SI (meters, seconds) before calculations to avoid errors.

When to use: Before solving numerical problems involving distance, time, velocity.

Tip: For relative velocity problems, choose a reference frame and apply vector addition/subtraction carefully.

When to use: When two or more objects move relative to each other.

Tip: Interpret velocity-time graphs by calculating area under the curve for displacement.

When to use: When given velocity-time graphs in problems.

Common Mistakes to Avoid

❌ Confusing distance with displacement and using them interchangeably.
✓ Remember distance is scalar and total path length; displacement is vector and shortest path.
Why: Students often ignore direction and treat displacement as distance.
❌ Ignoring direction when calculating velocity, leading to incorrect sign or magnitude.
✓ Always include direction and use vector notation for velocity.
Why: Velocity is a vector; neglecting direction changes the meaning.
❌ Using average speed formula for average velocity problems.
✓ Use displacement/time for average velocity, not total distance/time.
Why: Average velocity depends on displacement, not total distance.
❌ Not converting units properly, mixing km/h with m/s.
✓ Convert all units to SI before calculations (1 km/h = 5/18 m/s).
Why: Unit inconsistency leads to wrong answers.
❌ Misinterpreting velocity-time graphs and confusing slope with area.
✓ Slope of velocity-time graph gives acceleration; area under curve gives displacement.
Why: Students confuse graphical concepts.
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