In this question, position vectors are given relative to a fixed origin - Edexcel - A-Level Maths Mechanics - Question 1 - 2022 - Paper 1

The acceleration a of a particle is the derivative of the velocity with respect to time (t). Given:

$v = 3 extbf{i} - 6 extbf{j}$

differentiating, we have

a(t) = rac{d}{dt}(3 extbf{i} - 6 extbf{j}) = 0 extbf{i} + 0 extbf{j} = extbf{0}$$

Thus, the acceleration is:

$a(t) = 0 ext{ m/s}^2$

To find the position vector, we integrate the velocity with respect to time:

$r = extbf{r}_0 + ext{integrate}(v)$

Assuming the initial position vector at t=0 is ( extbf{r}_0) = (0 extbf{i}, 0 extbf{j}), we integrate:

$r(t) = \textbf{i}(3t) - \textbf{j}(6t) + \textbf{C}$

At t = 1 second:

$r(1) = (3 extbf{i} - 6 extbf{j}) + \textbf{C}$
To determine C, we rely on given position vector at t = 4 seconds being (i - 4j):

Using this, we solve backwards to find C when t=4. After solving, we obtain: $r(1) = (3 extbf{i} - 6 extbf{j}) + extbf{C}$.