At time t seconds, a particle P has velocity v m s⁻¹, where v = 3t²i - 2j, t > 0 (a) Find the acceleration of P at time t seconds, where t > 0 - Edexcel - A-Level Maths Mechanics - Question 5 - 2021 - Paper 1

To determine when P is moving in the direction of $extbf{i} - extbf{j}$, we must set the direction of the velocity vector equal to that.

The velocity vector is: $3t^2 extbf{i} - 2 extbf{j}$

This means we must find t such that: rac{3t^2}{-2} = rac{1}{-1}

Cross-multiplying: $3t^2 = 2$

Solving for t, t^2 = rac{2}{3}

Taking the square root gives:

oot{2}3}{3}$$

To find the expression for the position vector r, we need to integrate the velocity function with respect to time.

The velocity is: $v = 3t^2 extbf{i} - 2 extbf{j}$

Integrating gives: r = rac{3}{3}t^3 extbf{i} - 2t extbf{j} + C

Given the condition when t = 1, r = - extbf{j}, we substitute: $- extbf{j} = 1 extbf{i} - 2(1) extbf{j} + C$

This results in: $C = extbf{i} - extbf{j}$

Thus: $r = t^3 extbf{i} - 2t extbf{j} + extbf{i} - extbf{j}$

To find the speed, we first calculate the magnitude of the velocity vector:

Using the previous expression for v at any t:

Magnitude of v:

ightarrow ext{ Calculation: }$$ $$|v| = ext{√}((3t^2)^2 + (-2)^2) = ext{√}(9t^4 + 4)$$ Setting this equal to 10 and solving gives: $$ ext{√}(9t^4 + 4) = 10$$ Squaring both sides: $$9t^4 + 4 = 100$$ $$9t^4 = 96$$ $$t^4 = rac{96}{9} = rac{32}{3}$$ $$t = oot{4}{(rac{32}{3})}$$ To find r at this t, substitute back into the expression for r we derived above and calculate the distance using: $$ ext{distance} = ext{√}((x)^2 + (y)^2)$$.