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 $$ v = t^2 oldsymbol{i} - 4 oldsymbol{j}$$ When t = 1, P is at the point A and when t = 4, P is at the point B - Edexcel - A-Level Maths Mechanics - Question 6 - 2018 - Paper 1

We first find the position vectors for t = 1 and t = 4:

• For $t = 1$: $r(1) = \frac{1^3}{3} \boldsymbol{i} - 4(1) \boldsymbol{j} + C = \frac{1}{3} \boldsymbol{i} - 4 \boldsymbol{j} + C$
• For $t = 4$: $r(4) = \frac{4^3}{3} \boldsymbol{i} - 4(4) \boldsymbol{j} + C = \frac{64}{3} \boldsymbol{i} - 16 \boldsymbol{j} + C$

To find C, substitute the position of P when $t=1$ at point A.

To find the distance AB, we first calculate the position vectors:

Let point A be $\frac{1}{3} \boldsymbol{i} - 4 \boldsymbol{j}$ and point B be $\frac{64}{3} \boldsymbol{i} - 16 \boldsymbol{j}$.

The distance between A and B is given by:

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

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

Thus, the exact distance is: $AB = \sqrt{585} = \frac{3 \sqrt{65}}{2}$.