At time t seconds a particle, P, has position vector r metres, with respect to a fixed origin, such that $$r = (3t^2 - 5t) i + (8t - t^2) j$$ 14 (a) Find the exact speed of P when t = 2 14 (b) Bella claims that the magnitude of acceleration of P will never be zero - AQA - A-Level Maths Pure - Question 14 - 2020 - Paper 2

To find the speed of the particle P, we first need to differentiate the position vector r with respect to time t to get the velocity vector v.

Given, $r = (3t^2 - 5t) i + (8t - t^2) j$

Differentiating:
$v = \frac{dr}{dt} = \left( \frac{d}{dt}(3t^2 - 5t) \right)i + \left( \frac{d}{dt}(8t - t^2) \right)j$
Calculating the derivatives gives:

• For the i-component: $\frac{d}{dt}(3t^2 - 5t) = 6t - 5$
• For the j-component:
  $\frac{d}{dt}(8t - t^2) = 8 - 2t$
  Thus, the velocity vector v is: $v = (6t - 5)i + (8 - 2t)j$
  Now substituting t = 2: $v = (6(2) - 5)i + (8 - 2(2))j$
  Calculating each component: $v = (12 - 5)i + (8 - 4)j$
  $v = 7i + 4j$

To find the speed of the particle P, we calculate the magnitude of the velocity vector v:

$\text{Speed} = |v| = \sqrt{(7)^2 + (4)^2} = \sqrt{49 + 16} = \sqrt{65} \approx 8.06 m/s$