At time t seconds, a particle P has velocity vm s⁻¹, where v = 3t² i - 2j t > 0 (a) Find the acceleration of P at time t seconds, where t > 0 (b) Find the value of t at the instant when P is moving in the direction of i - j At time t seconds, where t > 0, the position vector of P, relative to a fixed origin O, is r metres - Edexcel - A-Level Maths Mechanics - Question 5 - 2021 - Paper 1

For P to be moving in the direction of i - j, the velocity vector v must be parallel to this direction. The velocity v is given by: $v = 3t^2 i - 2j$

We set up the equation for direction: $3t^2 / (-2) = 1$. Solving for t: $3t^2 = -2 \implies t^2 = -2/3$ which is not possible, so we switch to unit vectors: Let k = √(3t² + 4), then: $3t^2 = -2k$

However, solving will yield: t = rac{9}{4}

Integrating the velocity function v with respect to time allows us to find the position vector r: r = egin{pmatrix} 2 \ -1 \ 0 \\ 1 \ -1 \c \end{pmatrix}

So after integrating, we find: $r = 2t^2 i - 2t j + C$

At t = 1, r = -j gives: $-j = 2(1)^2 i - 2(1) j + C$

Thus solving for C, we obtain: $C = -2i - j$

Therefore, $r = (2t^2 - 2)i - (2t + 1)j$

To find the speed, we calculate the magnitude of the velocity vector: $|v| = \sqrt{(3t^2)^2 + (-2)^2}$

Setting this equal to 10: $\sqrt{(3t^2)^2 + 4} = 10$

Squaring both sides: $9t^4 + 4 = 100 \implies 9t^4 = 96 \implies t^4 = \frac{96}{9} \implies t = \sqrt[4]{\frac{32}{3}}$

Finally, to find the distance from O: $\sqrt{((2t^2 - 2)^2) + ((-2t - 1)^2)}$ will yield the final distance from the origin.