Question

question 11 (1 point)
a squirrel runs 15.0 m n and then turns and runs 8.0 m w. what is the squirrels displacement?
17 m w 28° n
23 m w 28° n
23 m w 62° n
17 m w 62° n

Explanation:

Step1: Calculate the magnitude of displacement

The squirrel's two - part motion forms a right - triangle. The two legs of the right - triangle are \(a = 15.0\space m\) (north) and \(b=8.0\space m\) (west).
Using the Pythagorean theorem \(d=\sqrt{a^{2}+b^{2}}\), where \(a = 15\space m\) and \(b = 8\space m\).

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Step2: Calculate the direction of displacement

Let \(\theta\) be the angle measured from the west direction to the displacement vector.
We know that \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\), where the opposite side to the angle \(\theta\) (with respect to the west - north right - triangle) is the northward displacement \(a = 15\space m\) and the adjacent side is the westward displacement \(b = 8\space m\).
\(\tan\theta=\frac{15}{8}=1.875\)
\(\theta=\tan^{- 1}(1.875)\approx62^{\circ}\)
The direction is \(W62^{\circ}N\)

Answer:

17 m [W \(62^{\circ}\) N]