A particle P of mass 2 kg is moving under the action of a constant force F newtons - Edexcel - A-Level Maths Mechanics - Question 3 - 2007 - Paper 1

Using Newton's second law, we know that the force (F) is given by:

$$F = m \mathbf{a},$$

where (m = 2 \text{ kg}) and (\mathbf{a} = 3\mathbf{i} - 1.5\mathbf{j}). Thus:

$$F = 2(3\mathbf{i} - 1.5\mathbf{j}) = 6\mathbf{i} - 3\mathbf{j}.$$

To find the magnitude of (F), we calculate:

$$|F| = \sqrt{(6)^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45} \approx 6.71 ext{ N}.$$

To find the velocity at t = 6 s, we can use the initial velocity and the acceleration:

$$\mathbf{v}_{6} = \mathbf{u} + \mathbf{a} \Delta t,$$

where (\Delta t = 6 \text{ s}) and using (\mathbf{u} = 3\mathbf{i} + 2\mathbf{j}) and (\mathbf{a} = 3\mathbf{i} - 1.5\mathbf{j}):

$$\mathbf{v}_{6} = (3\mathbf{i} + 2\mathbf{j}) + (3\mathbf{i} - 1.5\mathbf{j}) \cdot 6$$

Calculating this gives:

$$= (3 + 18)\mathbf{i} + (2 - 9)\mathbf{j} = 21\mathbf{i} - 7\mathbf{j}.$$

Thus, the velocity of P at t = 6 s is:

$$\mathbf{v}_{6} = 21\mathbf{i} - 7\mathbf{j} \text{ m s⁻¹}.$$