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

Firstly, we need to calculate the acceleration using Newton's second law, where the resultant force is acting on $P$:

$R = ma$

Here, the mass $m = 1.5\mathrm{kg}$. From part (a), we already determined:

$R = (6 + 5)i + (-2)j = 11i - 2j$

Setting this equal to $ma$ gives:

$11i - 2j = 1.5a$

Thus, the acceleration $a$ is:

$a = \left(\frac{11}{1.5}, \frac{-2}{1.5}\right) = (\frac{22}{3}, -\frac{4}{3})\mathrm{m \ s^{-2}}$

Now, we can solve for velocity at $t = 2$ seconds using:

$v = u + at$

Where, at $t=0$, $u = (-2i + 4j)$:

$v = (-2i + 4j) + 2 \left(\frac{22}{3}i - \frac{4}{3}j\right)$

Calculating gives:

$v = -2i + 4j + \left(\frac{44}{3}i - \frac{8}{3}j\right) = \left(-2 + \frac{44}{3}\right)i + \left(4 - \frac{8}{3}\right)j$

This simplifies to:

$v = \left(\frac{-6 + 44}{3}\right)i + \left(\frac{12 - 8}{3}\right)j = \left(\frac{38}{3} i + \frac{4}{3} j\right)$

Finally, we find the speed by calculating the magnitude:

$\text{Speed} = \sqrt{\left(\frac{38}{3}\right)^2 + \left(\frac{4}{3}\right)^2}$

Calculating yields:

$\text{Speed} = \frac{1}{3}\sqrt{1440} = \frac{1}{3}(12\sqrt{10}) = 10 \mathrm{m \ s^{-1}}$