[In this question i and j are horizontal unit vectors due east and due north respectively and position vectors are given relative to the fixed point O.] A particle P moves with constant acceleration - Edexcel - A-Level Maths Mechanics - Question 8 - 2018 - Paper 1

To determine the acceleration, we can use the formula for position:
$s = ut + \frac{1}{2}at^2$
At time t = 0, the particle is at point O with position vector at that time as:
$r(0) = 0$, and its velocity is given as:
$v(0) = (2i - 3j) \text{ ms}^{-1}$
For t = 2 seconds, the position vector given is (7i - 10j) m.
Substituting these values into the equation for position:
$(7i - 10j) = (2i - 3j)(2) + \frac{1}{2}a(2)^2$
Simplifying the right-hand side:
$\Rightarrow (7i - 10j) = (4i - 6j) + 2a$
Rearranging gives:
$7i - 10j - (4i - 6j) = 2a$
Thus,
$3i - 4j = 2a \Rightarrow a = \frac{3}{2}i - 2j \text{ ms}^{-2}$
Next, we calculate the magnitude of acceleration:
$|a| = \sqrt{\left(\frac{3}{2}\right)^2 + (-2)^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{9}{4} + \frac{16}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2} = 2.5 \text{ ms}^{-2}$
Therefore, the magnitude of the acceleration is confirmed to be 2.5 ms⁻².