A-level Maths revision · Mechanics

Motion in two dimensions

The idea

In two dimensions, position, velocity and acceleration are vectors. With constant velocity, r=r0+vt\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t. With variable motion, differentiate r\mathbf{r} component by component to get v\mathbf{v}, and again for a\mathbf{a}.

  1. Write the position as r=xi+yj\mathbf{r} = x\mathbf{i} + y\mathbf{j}.
  2. Differentiate or integrate each component separately.
  3. Speed is the magnitude of velocity: vx2+vy2\sqrt{v_x^2 + v_y^2}.

Watch out Include the starting position: r=r0+vt\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t.

See it worked

Worked example

r=(2+3t)i+(1−4t)j\mathbf{r} = (2 + 3t)\mathbf{i} + (1 - 4t)\mathbf{j}. Find the velocity and speed.

  1. v=drdt=3i−4j\mathbf{v} = \frac{d\mathbf{r}}{dt} = 3\mathbf{i} - 4\mathbf{j}.
  2. Speed =9+16= \sqrt{9 + 16}.
  3. =5 m s−1= 5\text{ m s}^{-1}.

Answer 3i−4j3\mathbf{i} - 4\mathbf{j}; speed 5 m s−15\text{ m s}^{-1}

Your turn

Question 1Non-calculator3 marks

A particle starts at (i+2j) m(\mathbf{i} + 2\mathbf{j})\text{ m} and moves with constant velocity (3i+4j) m s−1(3\mathbf{i} + 4\mathbf{j})\text{ m s}^{-1}. Find its position after 2 s2\text{ s}. Give the i\mathbf{i} and j\mathbf{j} components.

Question 2Non-calculator3 marks

Find the speed of a particle with velocity (6i−8j) m s−1(6\mathbf{i} - 8\mathbf{j})\text{ m s}^{-1}.

Question 3Non-calculator3 marks

r=t2i+(3t−1)j\mathbf{r} = t^2\mathbf{i} + (3t - 1)\mathbf{j}. Find the speed at t=2t = 2.

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