6. At time t seconds, where t > 0, a particle P moves in the x-y plane in such a way that its velocity v ms⁻¹ is given by $$ extbf{v} = t^2 extbf{i} - 4 extbf{j}$$ - Edexcel - A-Level Maths Mechanics - Question 6 - 2018 - Paper 1

At t = 1:

$\textbf{r}(1) = \frac{1^3}{3} \textbf{i} - 4(1) \textbf{j} + \textbf{C} = \frac{1}{3} \textbf{i} - 4 \textbf{j} + \textbf{C} = \textbf{A}$.

At t = 4:

$\textbf{r}(4) = \frac{4^3}{3} \textbf{i} - 4(4) \textbf{j} + \textbf{C} = \frac{64}{3} \textbf{i} - 16 \textbf{j} + \textbf{C} = \textbf{B}$.

The distance AB can be calculated using:

$AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2},$

Substituting the coordinates of points A and B:

$AB = \sqrt{\left(\frac{64}{3} - \frac{1}{3}\right)^2 + (-16 + 4)^2} = \sqrt{\left(\frac{63}{3}\right)^2 + (-12)^2} = \sqrt{21^2 + 12^2} = \sqrt{441 + 144} = \sqrt{585} = \sqrt{\frac{904}{226}}.$

Thus, the exact distance AB is:

$AB = \frac{\sqrt{904}}{\sqrt{226}}$.