Two particles, P and Q, move in the same horizontal plane - AQA - A-Level Maths Pure - Question 16 - 2022 - Paper 2

For particle P, which starts at position vector $(-4i + 5j)$ with constant acceleration $(3i - 4j)$, the position vector at time t seconds can be derived from the formula:

$ext{Position vector } extbf{r}_P = extbf{r}_0 + extbf{v}_0 t + \frac{1}{2} extbf{a} t^2$

Since it starts from rest, the initial velocity $extbf{v}_0 = 0$. Therefore, the expression simplifies to:

$extbf{r}_P = (-4i + 5j) + 0 + \frac{1}{2}(3i - 4j)t^2$

Consequently, $extbf{r}_P = (-4i + 5j) + \frac{3}{2}ti - 2tj^2$

Thus, the position vector of P is: $extbf{r}_P = \left( -4 + \frac{3}{2}t \right)i + \left( 5 - 2t \right)j.$

To prove that the paths of P and Q are not collinear, we can check the direction vectors derived earlier. From part (a), we know:

• Direction vector of particle P: $(3i - 4j)$
• Direction vector of particle Q: $(9i + (c + 1)j)$

Now substituting $c = -13$ in Q's direction vector gives: $(9i + (c + 1)j) = (9i + (-12)j) = 9i - 12j.$

To check collinearity, we set the ratios equal: $\frac{3}{9} = \frac{-4}{-12}.$

Simplifying gives: $\frac{1}{3} = \frac{1}{3}.$

While this shows proportionality, we must check the time equations of P and Q. The $t^2$ terms from both paths make them distinct as not all values of $t$ produce the same vectors for both paths as was shown in part (b). Therefore, paths are confirmed as not collinear.