Question

1. david walks 3 km north, then turns and walks 4 km east. what is his displacement? km at east of north

Explanation:

Step1: Analyze the displacement as a right - triangle

David's movements form a right - triangle. The northward walk (\(y = 3\) km) and eastward walk (\(x=4\) km) are the two legs of the right - triangle. The displacement \(d\) is the hypotenuse of the right - triangle.

Step2: Calculate the magnitude of displacement using the Pythagorean theorem

According to the Pythagorean theorem \(d=\sqrt{x^{2}+y^{2}}\). Substitute \(x = 4\) km and \(y = 3\) km into the formula:

$$
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Step3: Calculate the direction

Let \(\theta\) be the angle east of north. We know that \(\tan\theta=\frac{x}{y}\). Substitute \(x = 4\) km and \(y = 3\) km: \(\tan\theta=\frac{4}{3}\), then \(\theta=\arctan(\frac{4}{3})\approx53.1^{\circ}\)

Answer:

\(5\) km at \(53.1^{\circ}\) east of north