A particle P moves with acceleration (4i - 5j) ms² - Edexcel - A-Level Maths Mechanics - Question 2 - 2020 - Paper 1

To find T, we can use the displacement formula:

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

Where:

• $u = -2i + 2j$
• $a = 4i - 5j$
• $t = T$

Substituting the values:

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

At point A, the position vector is given as $(λi - 4.5j)$, thus: $(-2i + 2j)T + (2i - \frac{5}{2}j) T^2 = (λi - 4.5j)$

From this, we can equate components:

1. For the i component: $-2T + 2T^2 = λ$
2. For the j component: $2T - \frac{5}{2}T^2 = -4.5$

Rearranging the second equation gives us: $5T^2 - 4T - 9 = 0$

Using the quadratic formula where $a = 5$, $b = -4$, and $c = -9$: $T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Calculating: $T = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 5 \cdot (-9)}}{2 \cdot 5}$ $= \frac{4 \pm \sqrt{16 + 180}}{10}$ $= \frac{4 \pm \sqrt{196}}{10}$ $= \frac{4 \pm 14}{10}$

Thus:

1. $T = \frac{18}{10} = 1.8$ (valid as T > 0)
2. $T = \frac{-10}{10} = -1$ (not valid)

Hence, the value of T is 1.8.

Substituting the value of T into the equation for λ: $λ = -2T + 2T^2$

Replacing T with 1.8: $λ = -2(1.8) + 2(1.8^2)$

Calculating: $= -3.6 + 2(3.24)$ $= -3.6 + 6.48$ $= 2.88$

Thus, the value of λ is 2.88.