A car travels 10 km southeast and then 15 km in a direction 60° north of east. Find the magnitude of the car's resultant vector. (and the direction)

60×10 =600÷15=40

Step-by-step explanation:

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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]

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step-by-step explanation:

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