All subjects · Mathematics · WME01 (M1)

Vectors with constant acceleration and meeting problems

The suvat equations also work in vector form. Combined with position vectors, they let you decide whether two moving objects meet or how far apart they are.

Vector suvat

With constant acceleration a (a vector):

Acceleration = change in velocity ÷ time: a = (v - u)/t. Speed is |v| and the direction of motion is the direction of v.

Do they meet?

Write each position vector in terms of t. They meet if there is one value of t that makes both the i parts and the j parts equal. Solve one component equation, then check the other.

Distance apart

The vector from A to B at time t is r_B - r_A. Its magnitude is the distance between them. Questions may ask when they are a given distance apart, which leads to a quadratic in t.

Worked example

Position vectors are in km and t is in hours. Ship A: r = (2i + 10j) + t(3i - j). Ship B: r = (8i + 4j) + t(i + j). Do they meet?

  1. i parts: 2 + 3t = 8 + t gives t = 3
  2. j parts at t = 3: A has 10 - 3 = 7 and B has 4 + 3 = 7
  3. Both match

Answer: Yes, at t = 3 at the point 11i + 7j

Key idea: v = u + at and r = r_0 + ut + (1/2)at2 in vector form. To meet, both components must match at the same t.

Check yourself

  1. A particle has initial velocity (2i - 3j) m/s and constant acceleration (i + 2j) m/s2. Find its speed after 4 s, to 3 significant figures.
    Show answer

    7.81 m/s. v = (2 + 4)i + (-3 + 8)j = 6i + 5j, so speed = sqrt(61) = 7.81 m/s.

  2. A particle's velocity changes from (3i + j) m/s to (11i - 5j) m/s in 2 s. Find the magnitude of its acceleration.
    Show answer

    5 m/s2. a = (8i - 6j)/2 = 4i - 3j, magnitude 5 m/s2.

  3. Using the ships in the worked example, find the distance between them, in km, at t = 0, to 3 significant figures.
    Show answer

    8.49 km. r_B - r_A = 6i - 6j at t = 0, so the distance is sqrt(72) = 8.49 km.

  4. A particle has velocity v = (4 + t)i + (-6 + 2t)j m/s. Find the time when it is moving parallel to i.
    Show answer

    3 s. Parallel to i: j part zero, so -6 + 2t = 0 and t = 3.

  5. A particle starts from rest at O with acceleration (2i + 4j) m/s2. Find its distance from O after 3 s, to 3 significant figures.
    Show answer

    20.1 m. r = (1/2)(2i + 4j)(9) = 9i + 18j, so the distance is sqrt(405) = 20.1 m.

  6. To show two ships do NOT meet, a student finds that the i components are equal at t = 2. What must they check next?
    • Whether the ships have the same speed
    • Whether the i components are equal at t = 3
    • Whether the accelerations are equal
    • Whether the j components are also equal at t = 2
    Show answer

    Whether the j components are also equal at t = 2. Meeting needs both components equal at the same time.

  7. A particle has velocity (5i + 12j) m/s and the force on it gives acceleration (-i - 2.4j) m/s2. Find the time when it comes to rest.
    Show answer

    5 s. v = (5 - t)i + (12 - 2.4t)j. Both parts are zero when t = 5.

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Last updated 7 November 2026. Spotted a mistake? Tell us.