Two forces $(4i - 2j) \, N$ and $(2i + qj) \, N$ act on a particle $P$ of mass $1.5 \, kg$ - Edexcel - A-Level Maths Mechanics - Question 2 - 2014 - Paper 1

Using the initial velocity of the particle $P$, we have:

$v = (-2i + 4j) \, m \, s^{-1}$.

The acceleration, derived from the resultant force and mass, is given by:

$a = \frac{R}{m} = \frac{(6i + (q - 2)j)}{1.5} = (4i + (\frac{q - 2}{1.5})j)$.

When substituting $q = 6$, the acceleration becomes:

$a = (4i + 0j) = 4i \, m \, s^{-2}$.

To find the velocity at time $t = 2$ seconds, use:

$v = u + at$

where $u = (-2i + 4j)$ and $t = 2$ seconds. Thus:

$v = (-2i + 4j) + 2(4i)$

This results in:

$v = (-2 + 8)i + 4j = 6i + 4j \, m \, s^{-1}$.

To find the speed, compute the magnitude of the velocity vector:

$|v| = \sqrt{(6)^2 + (4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \, m \, s^{-1}$.

This approximates to 10 m/s.