Three particles \(A, B\) and \(C\), each of mass \(m\), are moving in a plane such that at time \(t\) their position vectors with respect to the origin \(O\) are \[\begin{align*} (2t+1)\mathbf{i}&+(2t+3)\mathbf{j} \\ (10-t)\mathbf{i}&+(12-t)\mathbf{j} \\ (3t^2-4t+1)\mathbf{i}&+ (-3t^2+2t)\mathbf{j} \end{align*}\]


  1. Show that the centre of mass of these three particles moves in a straight line and find the Cartesian equation of this line.

Since all the masses are equal, the centre of mass has the position vector that is the average of the three positions,

\[\dfrac{1}{3}\left((2t+1)\mathbf{i}+(2t+3)\mathbf{j} + (10-t)\mathbf{i}+(12-t)\mathbf{j}+ (3t^2-4t+1)\mathbf{i}+ (-3t^2+2t)\mathbf{j}\right)\] \[=(t^2-t+4)\mathbf{i} + (-t^2+t+5)\mathbf{j}.\]

We can find the equation of the locus of this point by putting \(x = t^2-t+4\), \(y = -t^2+t+5\) and then eliminating \(t\).

Adding, we find that \(x+y = 9\), which is a straight line.

Find also the value of \(t\) for which the centre of mass is instantaneously at rest.

The centre of mass is at rest when it’s \(x\)-coordinate is stationary. The \(y\)-coordinate is stationary at the same moment, since \(x+y = 9\). \(x\) is a quadratic function of \(t\) so we can complete the square: \[x = t^2-t+4 = (t-\tfrac{1}{2})^2+4-\tfrac{1}{4}.\] Hence the centre of mass is stationary at \(t=\dfrac{1}{2}\).

We could alternatively differentiate with respect to \(t\) to find the velocity.

  1. Verify that the particles \(A\) and \(B\) are both moving along the straight line with equation \(y=x+2\) and that they collide when \(t = 3\).

The position of \(A\) is \((2t+1)\mathbf{i}+(2t+3)\mathbf{j}\) which has locus \(x = 2t+1\), \(y = 2t+3\). Eliminating \(t\) we have \(y = x + 2\).

The position of \(B\) is \((10-t)\mathbf{i}+(12-t)\mathbf{j}\) which has locus \(x = 10-t\), \(y=12-t\). Eliminating \(t\) we have \(y=x+2\).

So \(A\) and \(B\) are both moving along this line.

They will collide when \(x = 2t+1 = 10-t\), that is when \(t=3\).