A particle P of mass 0.4 kg moves under the action of a single constant force F newtons - Edexcel - A-Level Maths Mechanics - Question 3 - 2008 - Paper 1

The acceleration vector of particle P is given as $(6i + 8j) \, \text{m s}^{-2}$. The mass of the particle is 0.4 kg. The force can be calculated using Newton's second law:

$F = ma = 0.4 \times (6i + 8j) = (2.4i + 3.2j) \, \text{N}.$

To find the magnitude of F, we calculate:

$|F| = \sqrt{(2.4)^{2} + (3.2)^{2}} = \sqrt{5.76 + 10.24} = \sqrt{16} = 4 \, \text{N}.$

To find the velocity at time t = 5 seconds, we will use the initial velocity and add the acceleration vector multiplied by time:

For $t = 0$, the initial velocity $v_0 = 9i - 10j$.

The acceleration is $6i + 8j$.

The velocity at time $t$ is:

$v = v_0 + at = (9i - 10j) + (6i + 8j) \cdot t.$

Thus, for $t = 5$:

$v = (9 + 30)i + (-10 + 40)j = 39i + 30j \, \text{m s}^{-1}.$