A ship travels 100 km on bearing 030°, then 100 km on bearing 150°. What is the magnitude and direction of the resulting displacement?

Prepare for the Navy Nationals UPI Test with essential study materials. Use flashcards and multiple-choice questions that provide hints and explanations. Ace your exam with confidence!

Multiple Choice

A ship travels 100 km on bearing 030°, then 100 km on bearing 150°. What is the magnitude and direction of the resulting displacement?

Explanation:
Think in terms of components: a bearing tells you how far to the east and how far to the north, so break each leg into east-west and north-south parts. For a distance d and bearing θ, the east component is d sin θ and the north component is d cos θ. First leg: 100 km at 30° gives east = 100 sin30 = 50 km, north = 100 cos30 ≈ 86.60 km. Second leg: 100 km at 150° gives east = 100 sin150 = 50 km, north = 100 cos150 ≈ -86.60 km. Add the components: total east = 50 + 50 = 100 km, total north ≈ 86.60 + (-86.60) = 0 km. So the displacement is 100 km to the east, i.e., bearing 090°. Magnitude is 100 km. The north-south components cancel, leaving a pure eastward displacement.

Think in terms of components: a bearing tells you how far to the east and how far to the north, so break each leg into east-west and north-south parts. For a distance d and bearing θ, the east component is d sin θ and the north component is d cos θ.

First leg: 100 km at 30° gives east = 100 sin30 = 50 km, north = 100 cos30 ≈ 86.60 km.

Second leg: 100 km at 150° gives east = 100 sin150 = 50 km, north = 100 cos150 ≈ -86.60 km.

Add the components: total east = 50 + 50 = 100 km, total north ≈ 86.60 + (-86.60) = 0 km. So the displacement is 100 km to the east, i.e., bearing 090°. Magnitude is 100 km. The north-south components cancel, leaving a pure eastward displacement.

More Multiple Choice Questions
Subscribe

Get the latest from Passetra

You can unsubscribe at any time. Read our privacy policy