A-level · Kinematics

Motion in two dimensions with vectorsExam-style questions

Why the a → v → r calculus chain works component-by-component in 2D, that speed is the magnitude of the velocity vector, and how "parallel to" and "perpendicular to" a direction become simple component conditions.

33 questions · A-level level · mark scheme and worked solution on each · no sign-up to try

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  1. 1
    In this question you must show all stages of your working.

    Solutions relying entirely on calculator technology are not acceptable.

    [In this question, i\mathbf{i} and j\mathbf{j} are horizontal perpendicular unit vectors.]

    A particle PP moves on a smooth horizontal plane. At time tt seconds, t⩾0t \geqslant 0, the position vector of PP relative to a fixed origin OO is r\mathbf{r} metres, where

    r=(2t2−5t)i+(t3−6t2+9t)j\mathbf{r} = (2t^2 - 5t)\mathbf{i} + (t^3 - 6t^2 + 9t)\mathbf{j}
    (a) Find the speed of PP at the instant when PP is moving in the direction of the vector i\mathbf{i}

    (5 marks)
    (b) Find the exact magnitude of the acceleration of PP when t=3.5t = 3.5

    (4 marks)
    (c) Show that PP never returns to OO

    (2 marks)
    Level 4 · 11 marksTry it
  2. 2
    [In this question i\mathbf{i} and j\mathbf{j} are horizontal unit vectors directed due east and due north respectively and position vectors are given relative to a fixed origin OO.]

    An oil platform at sea is modelled as a fixed point PP with position vector (13i+7j)(13\mathbf{i} + 7\mathbf{j}) km. Ships are not allowed to enter a circular safety zone of radius 22 km centred on PP.

    A cargo ship CC is moving with constant velocity (−12i+16j) km h−1(-12\mathbf{i} + 16\mathbf{j})\ \text{km h}^{-1}. At 05:30 the ship is at the point with position vector (31i−14j)(31\mathbf{i} - 14\mathbf{j}) km. The ship is modelled as a particle.
    (a) Find the speed of CC.

    (2 marks)
    (b) Find, in terms of i\mathbf{i}, j\mathbf{j} and tt, the position vector of CC at time tt hours after 05:30

    (2 marks)
    (c) (i) Find the value of tt for which the distance between CC and PP is least.

    (ii) Hence determine whether CC will enter the safety zone if it continues on its present course.

    (7 marks)
    Level 4 · 11 marksTry it
  3. 3
    [In this question i\mathbf{i} and j\mathbf{j} are horizontal unit vectors directed due east and due north respectively and position vectors are given relative to a fixed origin OO.]

    Two helicopters, AA and BB, are flying at the same constant height. Each helicopter flies in a straight line at a constant velocity, and the helicopters are modelled as particles moving in the same horizontal plane.

    Helicopter AA flies directly over a beacon with position vector (−20i+41j)(-20\mathbf{i} + 41\mathbf{j}) km at 09:40 and directly over a second beacon with position vector (52i+11j)(52\mathbf{i} + 11\mathbf{j}) km at 10:00

    The position vector of AA, tt hours after 09:40, is rA\mathbf{r}_A km.
    (a) Show that rA=(−20+216t)i+(41−90t)j\mathbf{r}_A = (-20 + 216t)\mathbf{i} + (41 - 90t)\mathbf{j}

    (4 marks)
    (b) Find the speed of AA in m s−1\text{m s}^{-1}

    (2 marks)

    Helicopter BB flies with constant velocity (−120i−50j) km h−1(-120\mathbf{i} - 50\mathbf{j})\ \text{km h}^{-1}. At 09:40, BB is at the point with position vector (136i+16j)(136\mathbf{i} + 16\mathbf{j}) km.

    The paths of the two helicopters cross at the point XX.
    (c) Find the position vector of XX.

    (5 marks)
    Level 4 · 11 marksTry it
  4. 4
    A particle PP moves in a horizontal plane. The velocity of PP at time tt seconds is v m s−1\mathbf{v}\ \text{m s}^{-1}, where v=(2t2−t−7) i+(t2−5t−2) j,t≥0.\mathbf{v} = (2t^2 - t - 7)\,\mathbf{i} + (t^2 - 5t - 2)\,\mathbf{j}, \qquad t \ge 0.
    (a) Find the initial speed of PP.

    (3 marks)
    (b) Find the value of tt for which the velocity of PP is parallel to the vector (i−j)(\mathbf{i} - \mathbf{j}).

    (2 marks)
    (c) Find, in terms of tt, the acceleration of PP.

    (2 marks)
    (d) Find the time at which PP is accelerating in a direction perpendicular to j\mathbf{j}.

    (2 marks)
    Level 4 · 9 marksTry it
  5. 5
    A particle moves in a plane with velocity v=(6t−4) i+3t2 j m s−1\mathbf{v} = (6t - 4)\,\mathbf{i} + 3t^2\,\mathbf{j}\ \text{m s}^{-1}. When t=0t = 0 its position vector relative to a fixed origin OO is (2i+j)(2\mathbf{i} + \mathbf{j}) m. Find
    (a) its position vector at time tt,
    (b) its position vector when t=2t = 2, and (c) its distance from OO when t=2t = 2.
    Level 4 · 5 marksTry it

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