At time t seconds a particle, P, has position vector r metres, with respect to a fixed origin, such that $$r = (3t^2 - 5t) extbf{i} + (8t - t^2) extbf{j}$$ 14 (a) Find the exact speed of P when t = 2 - AQA - A-Level Maths Pure - Question 14 - 2020 - Paper 2

The speed of the particle P is the magnitude of the velocity vector:

$\text{Speed} = |v| = \sqrt{(7)^2 + (4)^2} = \sqrt{49 + 16} = \sqrt{65}$

Thus, the exact speed of P when t = 2 is $\sqrt{65} \, \text{m s}^{-1}$.

To check Bella’s claim, we need to find the acceleration vector a by differentiating the velocity vector v:

$a = \frac{dv}{dt} = \left( \frac{d}{dt}(6t - 5) \right) \textbf{i} + \left( \frac{d}{dt}(8 - 2t) \right) \textbf{j}$

Calculating the derivatives gives:

$a = (6) \textbf{i} + (-2) \textbf{j}$

The magnitude of the acceleration is:

$|a| = \sqrt{(6)^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40}$

Thus, the magnitude of the acceleration is $\sqrt{40}$, which is constant and non-zero. Therefore, Bella's claim is correct as the magnitude will never be zero.