Vectors in mechanics

Using vectors for displacement, velocity, acceleration and force, and resultants.

[In this question $\mathbf{i}$ and $\mathbf{j}$ are unit vectors due east and due north, and distances are in km.]

At noon, ship $A$ is at the point with position vector $(2\mathbf{i} + 5\mathbf{j})$ and ship $B$ is at $(-4\mathbf{i} + 3\mathbf{j})$ . Ship $A$ moves with constant velocity $(3\mathbf{i} - \mathbf{j})\,\text{km h}^{-1}$ and ship $B$ moves with constant velocity $(5\mathbf{i} + \mathbf{j})\,\text{km h}^{-1}$ .

(a) Find the speed of ship $A$ .

(b) Find the time at which ship $B$ is due north of ship $A$ , and the distance between the ships at that time.

(a) Speed $= \sqrt{3^2 + (-1)^2} = 3.16\,\text{km h}^{-1}$ (1)

A particle of mass $0.5\,\text{kg}$ is acted on by two forces, $\mathbf{F}_1 = (4\mathbf{i} - 2\mathbf{j})\,\text{N}$ and , where and are constants. No other forces act. The particle accelerates at in the direction of the vector .

M1 Unit vector in direction $3\mathbf{i} + 4\mathbf{j}$ is $\tfrac{1}{5}(3\mathbf{i} + 4\mathbf{j})$ (1)
A1 Acceleration $= 10 \times \tfrac{1}{5}(3\mathbf{i} + 4\mathbf{j}) = (6\mathbf{i} + 8\mathbf{j})\,\text{m s}^{-2}$ (1)
M1 $\mathbf{F} = m\mathbf{a}$ : $(4 + a)\mathbf{i} + (b - 2)\mathbf{j} = 0.5(6\mathbf{i} + 8\mathbf{j}) = 3\mathbf{i} + 4\mathbf{j}$ (1)
A1 $a = -1$ (1)
A1 $b = 6$ (1)

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