An object is moving on a straight line path travelling 4kms towards East and then turn towards North and travels 5Km and then turns towards West and travels 3Kms. Calculate total distance travelled and displacement. please answer with explanation very urgent.
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Answered by
1
distance=4+5+3=12km
now for displacement...
(4km east)-(3km West)=1km east
so in the reference of displacement the body have travelled 1km east and 5km north
displacement=√[(1)²+(5)²] =√[1+25]=√26km
now for displacement...
(4km east)-(3km West)=1km east
so in the reference of displacement the body have travelled 1km east and 5km north
displacement=√[(1)²+(5)²] =√[1+25]=√26km
Answered by
0
Total distance = 4km + 5km + 3 km = 12km
Figure is shown in the photo
AP = BD - CD
AP = 4 - 3
AP = 1km
Apply Pythagoras thereom
PD² = AP²+AD²
PD² = (1)²+(5)²
PD² = 26
PD = √26
PD ≈ 5.09km
so displacement is 5.09 (it is a shortest path traveled by the object)
Figure is shown in the photo
AP = BD - CD
AP = 4 - 3
AP = 1km
Apply Pythagoras thereom
PD² = AP²+AD²
PD² = (1)²+(5)²
PD² = 26
PD = √26
PD ≈ 5.09km
so displacement is 5.09 (it is a shortest path traveled by the object)
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