Question

1. what is the total displacement of a trip of 50 km w followed by a trip of 100 km n? 4 marks

Explanation:

Identify the given vectors

We represent the two displacements as vectors:

• \(\vec{d}_1 = 50\text{ km [W]}\) (West, along the negative x-axis)
• \(\vec{d}_2 = 100\text{ km [N]}\) (North, along the positive y-axis)

Calculate the magnitude of total displacement

The two vectors are perpendicular. We use the Pythagorean theorem:

$$ d_{\text{total}} = \sqrt{d_1^2 + d_2^2} $$

$$ d_{\text{total}} = \sqrt{50^2 + 100^2} = \sqrt{2500 + 10000} = \sqrt{12500} \approx 111.8\text{ km} $$

Determine the direction of displacement

We find the angle \(\theta\) north of west:

$$ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{100}{50} = 2 $$

$$ \theta = \arctan(2) \approx 63.4^\circ $$

Thus, the direction is \([\text{W } 63.4^\circ \text{ N}]\), or equivalently \([\text{N } 26.6^\circ \text{ W}]\).

State the final displacement vector

Combining the magnitude and direction gives:

$$ \vec{d}_{\text{total}} \approx 112\text{ km } [\text{W } 63^\circ \text{ N}] \quad (\text{or } 1.1 \times 10^2\text{ km } [\text{W } 63^\circ \text{ N}] \text{ using two significant figures}) $$

Answer:

\(112\text{ km } [\text{W } 63^\circ \text{ N}]\) (or \(1.1 \times 10^2\text{ km } [\text{W } 63^\circ \text{ N}]\) to two significant figures)