Example 1.8 Taking a Hike A hiker begins a trip by first walking 24.0 km southea

Question

Example 1.8 Taking a Hike A hiker begins a trip by first walking 24.0 km southeast from her car. She stops and sets up her tent for the night. On the second day, she walks 43.0 km in a direction 60.0° north of east, at which point she discovers a forest ranger's tower. y (km) (A) Determine the components of the hiker's displacement for each day lower (B) Determine the components of the hiker's resultant displacement R for the trip. Find an expression for R in terms of unit vectors x (km) Car 45.0° 60.0 SOLVE IT Tent The total displacement of the hiker is the vector Conceptualize We conceptualize the problem by drawing a sketch as in the figure. If we denote the displacement vectors on the first and second days by A and EB respectively, and use the car as the origin of coordinates we obtain the vectors shown in the figure R- A +B Categorize Drawing the resultant R, we can now categorize this problem as one we've solved before: an addition of two vectors. You should now have a hint of the power of categorization in that many new problems are very similar to problems we have already solved if we are careful to conceptualize them Once we have drawn the displacement vectors and categorized the problem, this problem is no longer about a hiker, a walk, a car, a tent, or a tower. It is a problem about vector addition, one that we have already solved Analyze Displacement A has a magnitude of 24.0 km and is directed 45.0° below the positive x axis

Explanation / Answer

Here ,

for the displacements

d1 = 20 * (cos(45 degree) i - j * sin(45 degree))

d2 = 32 * (cos(60) i + j * sin(60))

for the net displacement

Dnet= d1 + d2

Dnet = 20 * (cos(45 degree) i - j * sin(45 degree)) + 32 * (cos(60) i + j * sin(60))

Dnet= 30.1 i + 10.7 j km

magnitude = sqrt(30.1^2 + 10.7^2)

magnitude = 32 km

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direction = atan(10.7/30.1)

direction = 19.6 degree

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