A hiker makes four straight-line walks A 33 km at 48 0 B 19 km at 147 0 C 14 km

Question

A hiker makes four straight-line walks
A 33 km at 480

B 19 km at 1470

C 14 km at 970

D 26 km at 253

in random directions and lengths starting at position (41km, 41km), How far from the starting point is the hiker after these four legs of the hike? All angles are measured in a counter-clockwise direction from the positive x-axis. Answer in units of km

Explanation / Answer

If we assume all four walks of hiker four different vector. The final distance can be seen as the magnitude of vector-sum. Let hiker be at any point P(41km, 41km) and then 4 walks P->A->B->C->D final distance is magnitude of lenth DP PA=33(cos48 i +sin48 j) AB=19(cos147 i +sin147 j) BC=14(cos97 i +sin97 j) CD=26(cos253 i +sin253 j) (here i and j are the unit vector along x-axis and y-axis respectively) Now DP=PA+AB+BC+CD .....this is vector sum After calculating and simplifying, you get DP = -3.16i + 23.90j Magnitude of DP = sqrt[(-3.16)^2+(23.90)^2] = 24.11 kms.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.