Question

A car travels 20.0 km due north and then 95.0 km in a direction 60.0° west of north, as in Figure 1.15. Find the magnitude and direction of a single vector that gives the net effect of the car's trip. This vector is called the car's resultant disblacement.

Answer 1

R = 95 km, 30° west of north

Step-by-step explanation:

You didn't post a figure, but I think I can picture it.

60 west of north = 30 north of west

sin30 = y/95

y = sin30(95) = 47.5

cos30 = x/95

x = cos30(95) = 82.3

∑y = 20 + 47.5 = 67.5

∑x = 82.3

R² = 82.3² + 47.5² = 9029.54

R = √9029.54 = 95 (magnitude of the Resultant vector)

∅ = tan⁻¹ 47.5/82.3 = 30 north of west