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lizzyhearts
06.12.2021 •
Mathematics
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)
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Ответ:
60×10 =600÷15=40
Step-by-step explanation:
oh that's my answer
Ответ:
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]
Ответ: