This tutorial explains the difference between displacement and distance with visual examples and step-by-step calculations. Both concepts are fundamental in kinematics, the study of motion. More practice problems and detailed solutions are available on this site.
Distance is a scalar quantity (magnitude only) that represents the total length of the path traveled by a moving object, regardless of direction. It's always positive and measured in units of length (centimeters, meters, kilometers, etc.). The distance between two points depends on the path taken.
Displacement is a vector quantity (both magnitude and direction) that represents the straight-line change in position from initial to final point. It's the shortest distance between these two points with a specific direction. Displacement can be positive, negative, or zero, depending on the coordinate system. Mathematically, displacement \(\Delta \vec{x}\) is defined as:
\[ \Delta \vec{x} = \vec{x}_f - \vec{x}_i \]
where \(\vec{x}_f\) is the final position and \(\vec{x}_i\) is the initial position.
An object moves from point A to D along the red path shown below.
a) Find the total distance covered by the object.
b) Find the displacement of the object.
Solution:
a) Using the given scale (1 km per division), the total distance \(d\) is:
\[ d = AB + BC + CD = 2\text{ km} + 5\text{ km} + 2\text{ km} = 9\text{ km} \]
b) The displacement is the straight-line vector from initial position A to final position D. Its magnitude is:
\[ |\Delta \vec{x}| = AD = 5\text{ km} \]
The direction is eastward (to the right in the diagram).
An object moves from point A to B to C and then back to B as shown below.
a) Find the total distance covered.
b) Find the magnitude of the displacement.
Solution:
a) The total distance \(d\) is the sum of all path segments:
\[ d = AB + BC + CB = 5\text{ km} + 4\text{ km} + 4\text{ km} = 13\text{ km} \]
b) The initial position is A and final position is B, so the displacement magnitude is:
\[ |\Delta \vec{x}| = AB = 5\text{ km} \]
The displacement direction is from A to B (eastward).
An object moves from point A to B to C to D and back to A along the rectangle below.
a) Find the total distance covered.
b) Find the displacement.
Solution:
a) The total distance equals the rectangle's perimeter. Using the scale:
\[ d = 2 \times AB + 2 \times BC = 2 \times 5\text{ km} + 2 \times 3\text{ km} = 16\text{ km} \]
b) Since the object returns to its starting point, the initial and final positions are identical:
\[ \Delta \vec{x} = \vec{x}_f - \vec{x}_i = \vec{0} \]
The displacement is zero, demonstrating that distance can be nonzero even when displacement is zero.
An object moves from point A to B to C along the semicircle shown below (radius = 3 km).
a) Find the total distance covered.
b) Find the displacement.
Solution:
a) The path length is half the circumference of a circle with radius \(r = 3\text{ km}\):
\[ d = \frac{1}{2} \times 2\pi r = \pi r = 3\pi\text{ km} \approx 9.42\text{ km} \]
b) The displacement is the straight line from A to C, which is the circle's diameter:
\[ |\Delta \vec{x}| = 2r = 6\text{ km} \]
The direction is eastward (from leftmost to rightmost point).
An object moves from point A to B along the quarter-circle shown below (radius = 3 km).
a) Find the total distance covered.
b) Find the magnitude of the displacement.
Solution:
a) The path is one-quarter of a circle's circumference:
\[ d = \frac{1}{4} \times 2\pi r = \frac{\pi r}{2} = \frac{3\pi}{2}\text{ km} \approx 4.71\text{ km} \]
b) The displacement magnitude is the straight-line distance from A to B. Points A and B are at the ends of a quarter-circle, forming a right triangle with sides equal to the radius:
Using the Pythagorean theorem:
\[ AB^2 = r^2 + r^2 = 3^2 + 3^2 = 18 \]
\[ |\Delta \vec{x}| = AB = \sqrt{18} = 3\sqrt{2}\text{ km} \approx 4.24\text{ km} \]
Note that \(|\Delta \vec{x}| < d\), consistent with displacement magnitude being less than or equal to distance traveled.
Distance is relevant when considering energy expenditure, fuel consumption, or wear and tear, as these depend on the total path length. Displacement is crucial for analyzing net motion, velocity, and acceleration, as these describe changes in position regardless of the path taken.