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

Now substituting $t = 2$ into the velocity vector:

$extbf{v} = (6(2) - 5) extbf{i} + (8 - 2(2)) extbf{j} = (12 - 5) extbf{i} + (8 - 4) extbf{j} = 7 extbf{i} + 4 extbf{j}.$

The speed of $P$ is given by the magnitude of the velocity vector:

$ext{Speed} = | extbf{v}| = ext{sqrt}( (7)^2 + (4)^2 ) = ext{sqrt}(49 + 16) = ext{sqrt}(65) = 4 ext{.}1231 ext{ m/s}.$

To find the acceleration $extbf{a}$, differentiate the velocity vector $extbf{v}$ with respect to time $t$:

extbf{a} = rac{d extbf{v}}{dt} = (6) extbf{i} + (-2) extbf{j}.

The magnitude is given by:

Since the magnitude of the acceleration vector $| extbf{a}|$ evaluates to a positive value, Bella's claim that the magnitude of acceleration of $P$ will never be zero is correct. It holds for all values of $t$.