A particle P moves with constant acceleration (2i - 3j) ms² At time t = 0, P is moving with velocity 4i ms⁻¹ (a) Find the velocity of P at time t = 2 seconds - Edexcel - A-Level Maths Mechanics - Question 1 - 2021 - Paper 1

To find the position vector of particle P at time t = 3 seconds, we start with the kinematic equation for position:

$r = r_0 + ut + \frac{1}{2}at^2$

Where:

• $r_0$ is the initial position vector, given as $(i + j)$ m.
• $u$ is the initial velocity, which we calculated as $8i - 6j$ ms⁻¹ at t = 2 seconds.
• $a$ is the acceleration, which is $(2i - 3j)$ ms².
• $t$ is the time, which we will consider as 3 seconds.

Now we will substitute the values:

$r = (i + j) + (8i - 6j) imes 3 + \frac{1}{2}(2i - 3j)(3^2)$

Calculating step by step:

1. Determine $ut$:

   • $= (8i - 6j) imes 3 = 24i - 18j$

2. Determine $\frac{1}{2}at^2$:

   • $= \frac{1}{2}(2i - 3j) imes 9 = (1i - 1.5j) imes 9 = 9i - 13.5j$

Now adding up:

$r = (i + j) + (24i - 18j) + (9i - 13.5j)$

Combine like terms: $r = (1 + 24 + 9)i + (1 - 18 - 13.5)j$ $r = 34i - 30.5j$

Thus, the position vector of P relative to O at time t = 3 seconds is:

$r = 34i - 30.5j ext{ m}$