Statement of the given problem,
A car travels 20 km due north and then 35 km in a direction 60° west of north. What is the magnitude and direction of the car’s resultant displacement?
Let
î & ĵ denote two unit vectors along East & North directions respectively.
D denotes the resultant displacement vector of the given car.
Hence from above data we get as follows,
D = 20*ĵ - 35*(sin 60°)*î + 35*(cos 60°)*ĵ
or D = - 35*(sin 60°)*î + [20 + 35*(cos 60°)]*ĵ
Therefore,
the required magnitude of the car’s resultant displacement
= | D |
= √[{- 35*(sin 60°)}^2 + {20 + 35*(cos 60°)}^2]
= √[(35*√3/2)^2 + (20 + 35/2)^2]
= √2325 = 48.218 (km) [Ans]
the required direction of the car’s resultant displacement
= arctan [{20 + 35*(cos 60°)}/{- 35*(sin 60°)}]
= arctan [(20 + 35/2)/(- 35*√3/2)]
= arctan (-1.237)
= 51.05° (North of West)
or (90° - 51.05° =) 38.95° (West of North) [Ans]
