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Learn9MA0Paper 3 · Statistics and MechanicsKinematicsPosition, velocity and acceleration vectors
Kinematics · Subtopic 5 of 5

Position, Velocity and Acceleration Vectors

An object starting at r0\mathbf r_0 and moving with constant velocity v\mathbf v is at r=r0+vt\mathbf r=\mathbf r_0+\mathbf vt after time tt. Speed is the magnitude of velocity. "Due north" means the i\mathbf i-component is zero; two objects meet when their position vectors are equal at the same time.

Edexcel YMA01 WME01Edexcel 9MA0 Paper 3AQA 7357 Paper 2≈ 35 minRevision 5 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Use r=r0+vt\mathbf r=\mathbf r_0+\mathbf vt.
  • Find velocity from two positions, and speed from velocity.
  • Use v=u+at\mathbf v=\mathbf u+\mathbf at with vectors.
  • Find when an object is due north, east, etc., or when two objects meet.
Pause and think

A boat starts at (2i+j)(2\mathbf i+\mathbf j) km and each hour moves (3i−2j)(3\mathbf i-2\mathbf j) km. Where is it after 11, 22 and tt hours?

Section 2 of 14

Core idea

Start hereWill the boats collide?

Two boats move across a bay in straight lines. Where is each one at 3 pm, and how far apart are they? Writing positions as vectors that depend on time turns this into simple algebra.

The big idea

Constant velocity: r=r0+vt\mathbf r=\mathbf r_0+\mathbf vt. Constant acceleration: v=u+at\mathbf v=\mathbf u+\mathbf at, r=r0+ut+12at2\mathbf r=\mathbf r_0+\mathbf ut+\frac12\mathbf at^2.

1. Position from velocity

A boat starting at (2, 1) km with velocity (3i + 2j) km/h
24681012142468100east (km)north (km)rt = 0
r=(2i+j)+t(3i+2j)\mathbf r=(2\mathbf i+\mathbf j)+t(3\mathbf i+2\mathbf j). After 33 hours it is at (11,7)(11, 7). Its speed is ∣3i+2j∣=13=3.61|3\mathbf i+2\mathbf j|=\sqrt{13}=3.61 km/h.
r=r0+vt\mathbf r=\mathbf r_0+\mathbf vt
where
r0\mathbf r_0
position at t=0t=0.
v\mathbf v
constant velocity.
Worked exampleBoat A

At (2i+j)(2\mathbf i+\mathbf j) km, velocity 3i+2j3\mathbf i+2\mathbf j km h−1^{-1}. Position at time tt and t=3t=3; speed.

Goal
r\mathbf r; speed.
Strategy
Formula.
  1. r=(2+3t)i+(1+2t)j\mathbf r=(2+3t)\mathbf i+(1+2t)\mathbf j
    Collect components.
  2. t=3: 11i+7jt=3:\ 11\mathbf i+7\mathbf j
    Substitute.
  3. ∣v∣=13=3.61 km h−1|\mathbf v|=\sqrt{13}=3.61\text{ km h}^{-1}
    Speed is the magnitude of velocity.
Answer11i+7j11\mathbf i+7\mathbf j km; 3.613.61 km h−1^{-1}.
✓ CheckIn 3 h it moves 3×3.61=10.83\times3.61=10.8 km: from (2,1)(2,1) to (11,7)(11,7) is 81+36=10.8\sqrt{81+36}=10.8 ✓.
Worked exampleDistance between boats

Boat B: r=(10−t)i+(−1+4t)j\mathbf r=(10-t)\mathbf i+(-1+4t)\mathbf j. Distance apart at t=2t=2?

Goal
Distance.
Strategy
Subtract positions.
  1. A:(8,5),B:(8,7)A:(8,5),\quad B:(8,7)
    Substitute t=2t=2.
  2. AB→=2j⇒2 km\overrightarrow{AB}=2\mathbf j\Rightarrow2\text{ km}
    Magnitude.
Answer22 km.
✓ CheckSame xx-coordinate, so the gap is purely north–south ✓.

2. Constant acceleration

Worked exampleFinding acceleration

Velocity changes from i+2j\mathbf i+2\mathbf j to 7i−6j7\mathbf i-6\mathbf j in 4 s.

Goal
a\mathbf a, ∣a∣|\mathbf a|.
Strategy
a=v−ut\mathbf a=\frac{\mathbf v-\mathbf u}t.
  1. a=6i−8j4=1.5i−2j\mathbf a=\frac{6\mathbf i-8\mathbf j}4=1.5\mathbf i-2\mathbf j
    Subtract, divide.
  2. ∣a∣=2.25+4=2.5|\mathbf a|=\sqrt{2.25+4}=2.5
    Magnitude.
Answer1.5i−2j1.5\mathbf i-2\mathbf j; 2.52.5 m s−2^{-2}.
✓ Check∣6i−8j∣=10|6\mathbf i-8\mathbf j|=10; 10/4=2.510/4=2.5 ✓.
Common misconceptionPosition vs displacement r\mathbf r is measured from the origin; displacement is the change in r\mathbf r.
Pause and think

When is v=(4−t)i+3j\mathbf v=(4-t)\mathbf i+3\mathbf j parallel to j\mathbf j?

Summary

  • r=r0+vt\mathbf r=\mathbf r_0+\mathbf vt.
  • Vector suvat for constant a\mathbf a.
Key words
Position vector
Location relative to the origin.
HistoryNavigation by dead reckoning. Before satellites, sailors estimated position by "dead reckoning": starting point plus speed × time in a known direction, exactly r=r0+vt\mathbf r=\mathbf r_0+\mathbf vt. Christopher Columbus crossed the Atlantic in 1492 largely this way.
Fun factCollision courses. Air-traffic controllers' software continually computes when two aircraft would be closest, using their position and velocity vectors. If the closest approach is too near, it raises an alert long before any risk.
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