A particle P moves with acceleration (4i - 5j) ms² At time t = 0, P is moving with velocity (-2i + 2j) ms⁻¹ (a) Find the velocity of P at time t = 2 seconds - Edexcel - A-Level Maths Mechanics - Question 2 - 2020 - Paper 1

We first need to determine the position of particle P at any time t. The displacement can be calculated using:

$r = ut + \frac{1}{2} a t^2$

Substituting the values into the formula:

$r(t) = (-2i + 2j)t + \frac{1}{2}(4i - 5j)t^2$

This simplifies to:

$r(t) = (-2t + 2t + 2t^2)i + (2t - \frac{5}{2}t^2)j$

At time t = T, we know that:

• The position vector of A, $r(T) = (λi - 4.5j)$

Setting components equal:

For the i-component:

$\implies -2T + 2T + 2T^2 = λ$

For the j-component:

$\implies 2T - \frac{5}{2}T^2 = -4.5$

Rearranging,

$\frac{5}{2}T^2 - 2T - 4.5 = 0$

Using the quadratic formula $T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = \frac{5}{2}, b = -2, c = -4.5$:

This yields T ≈ 1.8 seconds.

Substituting T = 1.8 into the equation for λ from the i-component:

$λ = -2(1.8) + 2(1.8) + 2(1.8^2)$

Calculating:

$λ = -3.6 + 3.6 + 2(3.24)$

So, $λ = 6.48$

Thus, the value of λ is approximately 6.48.