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Motion in Two Dimensions · Lesson 1 of 6

Position and Displacement in a Plane

12 min read1 idea3 checks2 figures to playBy Asheshanand Acharya
What you will learn 0 / 1
Position and displacement as vectors
Maths you need
Look around
Where is the fielder?

A commentator says that the fielder is thirty metres from the batsman. That is not enough to place the fielder. Thirty metres in which direction? On a cricket ground, one distance never fixes a position. Two numbers are needed, or a distance and a direction. The last chapter needed only one number because the motion was along a line. A plane needs two.

you will be able to give the position of any point on a ground with two numbers and one origin, and to say what the fielder's displacement was without knowing the path the fielder ran.

Two axes, two numbers

Choose an origin O on the plane. Draw two perpendicular axes through O, the \(x\) axis and the \(y\) axis. Any point in the plane is then fixed by two numbers, its \(x\) coordinate and its \(y\) coordinate. The point \((3, 4)\,\text{m}\) lies \(3\,\text{m}\) along the \(x\) axis and \(4\,\text{m}\) along the \(y\) axis from O. The two numbers together are the position of the point.

The position can also be written as one arrow. The position vector \(\vec r\) is the arrow from the origin O to the point. Its length is the distance of the point from O. Its direction is the direction of the point as seen from O. In terms of the unit vectors \(\hat i\) along \(x\) and \(\hat j\) along \(y\),

\[ \vec r = x\,\hat i + y\,\hat j, \qquad |\vec r| = \sqrt{x^2 + y^2} \]

For the point \((3, 4)\,\text{m}\), \(\vec r = 3\hat i + 4\hat j\) and \(|\vec r| = \sqrt{9 + 16} = 5\,\text{m}\).

The position vector of a particle moving in a planeA particle moves along a curved path in a plane with x and y axes. A brown arrow from the origin to the particle is the position vector. Two readings give the x and y coordinates as the particle moves.123456789101234Ox (m)y (m) r P x = 1.0 m, y = 1.0 m
Fig 1.1Press play. A particle moves along a curved path in the plane. The brown arrow is its position vector \(\vec r\), always drawn from the origin O. The readings give the two components \(x\) and \(y\), which change together as the particle moves.
definition 3.1
Position vector

The position vector of a body is the vector from the origin of the frame of reference to the position of the body. In a plane, \(\vec r = x\,\hat i + y\,\hat j\), where \(x\) and \(y\) are the coordinates of the body.

Displacement is a vector in the plane

Suppose the body moves from position \(\vec r_1 = x_1\hat i + y_1\hat j\) to position \(\vec r_2 = x_2\hat i + y_2\hat j\). The displacement is the change in the position vector.

\[ \Delta\vec r = \vec r_2 - \vec r_1 = (x_2 - x_1)\,\hat i + (y_2 - y_1)\,\hat j \]

Displacement is found by subtracting the coordinates one axis at a time. The \(x\) component of the displacement is the change in \(x\). The \(y\) component is the change in \(y\). The path between the two positions does not enter at all. As in one dimension, displacement depends only on the end points.

Displacement as the difference of two position vectorsTwo position vectors r1 and r2 from the origin to two points on a curved path, and the displacement vector joining the tip of r1 to the tip of r2. Chips move the second point along the path; the displacement arrow follows the tips.123456789101234Ox (m)y (m) r₁ r₂ Δr 1 2 Δr = (4.0, 0.6) m, |Δr| = 4.0 m
Fig 1.2Two position vectors \(\vec r_1\) and \(\vec r_2\) and the displacement \(\Delta\vec r\) joining their tips. The grey curve is the path. Use the chips to move the second position: the displacement arrow follows the tips, never the curve.
RememberDisplacement in a plane
\[ \Delta\vec r = (x_2 - x_1)\,\hat i + (y_2 - y_1)\,\hat j, \qquad |\Delta\vec r| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Holds whenany motion in a plane, whatever the path between the two positions.
Symbols\((x_1, y_1)\) the initial position, \((x_2, y_2)\) the final position, \(\Delta\vec r\) the displacement vector
Unitsall in m
ExampleThe fielder's displacement

A fielder starts at the point \((10, 5)\,\text{m}\) and, after running along a curve to collect the ball, ends at \((22, 14)\,\text{m}\). Find the displacement vector and its magnitude.

Try it on paper first. The solution is one tap away when you are ready.
Checkpoint · numericUnanswered

A body moves from the point \((2, 1)\,\text{m}\) to the point \((5, 5)\,\text{m}\). Find the magnitude of its displacement in metres.

m± 1 %
Common mistake

The magnitude of the displacement is not \(\Delta x + \Delta y\). The two components are perpendicular, so they add like the two sides of a right triangle, by Pythagoras. \(3\,\text{m}\) east and \(4\,\text{m}\) north give \(5\,\text{m}\), not \(7\,\text{m}\).

CheckpointUnanswered

A runner completes one full lap of a \(400\,\text{m}\) athletics track and returns to the starting line. Which statement is correct?

Property
Position vector
Displacement vector
Drawn from
the origin O
the initial position
Drawn to
the body
the final position
Depends on the origin
yes
no
Depends on the path
no
no
Deeper

The displacement does not depend on where the origin is placed. Move the origin, and both \(\vec r_1\) and \(\vec r_2\) change by the same vector, so their difference \(\Delta\vec r\) does not change. This is why displacement, not position, is the quantity that velocity is built from.

Checkpoint · in your own wordsUnanswered

Why does a position in a plane need two numbers, while a position on a straight line needs only one?

Exam traps
  1. Adding the components arithmetically to get the magnitude of a vector.
    Components along perpendicular axes combine by Pythagoras: \(|\vec r| = \sqrt{x^2 + y^2}\).
  2. Confusing the position vector with the displacement vector.
    Position is measured from the origin. Displacement is measured from the previous position and does not depend on the origin.