Differentiating position vectors

Component by component

If r = f(t)i + g(t)j then

• v = dr/dt = f'(t)i + g'(t)j
• a = dv/dt = f''(t)i + g''(t)j.

Speed is |v|, the magnitude of the velocity. Pearson's examiners report candidates who substitute t into r instead of differentiating first, and others who give a vector when the question asks for a speed.

Directions

• Moving parallel to i: the j part of v is zero.
• Moving parallel to j (due north if j is north): the i part of v is zero (and the j part positive for north).
• Moving in the direction of pi + qj: v is a positive multiple of pi + qj, so the ratio of its parts is p : q.

Force from position

Newton's second law in vector form, F = ma, gives the resultant force once you have a. Its magnitude is m|a|.

Worked example

A particle has position vector r = (t³ - 3t)i + 4t² j metres. Find its speed and acceleration when t = 2.

1. v = (3t² - 3)i + 8t j; at t = 2, v = 9i + 16j.
2. Speed = sqrt(81 + 256) = 18.4 m/s.
3. a = 6t i + 8j; at t = 2, a = (12i + 8j) m/s².

Answer: Speed 18.4 m/s, a = (12i + 8j) m/s²

Check yourself

1. A particle has r = 2t² i + (t³ - 12t)j metres. Find its speed when t = 1, to 3 significant figures.
   Show answer

   9.85 m/s. v = 4t i + (3t² - 12)j = 4i - 9j at t = 1, so speed = sqrt(97) = 9.85 m/s.

2. For the same particle (r = 2t² i + (t³ - 12t)j), find the time t > 0 when it is moving parallel to i.
   Show answer

   2 s. j part of v: 3t² - 12 = 0, so t = 2.

3. For the same particle, find the magnitude of its acceleration when t = 2, to 3 significant figures.
   Show answer

   12.6 m/s². a = 4i + 6t j = 4i + 12j at t = 2, so |a| = sqrt(160) = 12.6 m/s².

4. The particle in the previous question has mass 0.5 kg. Find the magnitude of the resultant force on it when t = 2, to 3 significant figures.
   Show answer

   6.32 N. F = ma, so |F| = 0.5 × 12.6 = 6.32 N.

5. A particle has r = t² i + 5t j metres. Find the angle between its direction of motion and i when t = 2, in degrees to 1 decimal place.
   Show answer

   51.3 °. v = 2t i + 5j = 4i + 5j, so tan θ = 5/4 and θ = 51.3°.

6. A particle has r = (t² + 4t)i + t³ j metres. Find the time t > 0 when it is moving in the direction of i + j, to 3 significant figures.
   Show answer

   1.54 s. v = (2t + 4)i + 3t² j. Equal parts: 3t² = 2t + 4, so 3t² - 2t - 4 = 0 and t = 1.54 s.

7. A question gives r in terms of t and asks for the speed at t = 4. What is the first step?
   • Substitute t = 4 into r
   • Integrate r with respect to t
   • Differentiate r with respect to t
   • Find the magnitude of r
   Show answer

   Differentiate r with respect to t. Velocity is dr/dt; find it before substituting.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3), Unit M2
• Pearson Edexcel IAL Mechanics M2 (WME02) question paper, June 2025: front-cover instructions on g and accuracy
• Pearson Edexcel IAL Mechanics M2 (WME02) examiners' report, June 2025

Last updated 8 November 2026. Spotted a mistake? Tell us.