Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 1 (Q.No. 3)
3.
At a given instant ship A is travelling at 6 km/h due east and ship B is travelling at 8 km/h due north. The velocity of B relative to A is
Discussion:
22 comments Page 1 of 3.
Nabin Acharya said:
2 years ago
@All.
Don't get confused between resultant and relative velocity they are completely different (although in this particular case, the value is the same). The result can be easily determined from vector operation.
Here, Va = 6i + 0j and Vb = 0i + 8j, Thus Vba = Vb - Va = -6i + 8j,
Hence Magnitude of Vba = Sqrt (6^2 + 8^2) = 10 kmph.
Don't get confused between resultant and relative velocity they are completely different (although in this particular case, the value is the same). The result can be easily determined from vector operation.
Here, Va = 6i + 0j and Vb = 0i + 8j, Thus Vba = Vb - Va = -6i + 8j,
Hence Magnitude of Vba = Sqrt (6^2 + 8^2) = 10 kmph.
(1)
Onkar said:
3 years ago
Va = 6.
Vb = 8.
Va/b = √(va+vb)= √6^2 + 8^2 = √100.
Va/b = 10.
Vb = 8.
Va/b = √(va+vb)= √6^2 + 8^2 = √100.
Va/b = 10.
(2)
Larry L said:
4 years ago
Va = 6kmph.
Vb = 8kmph.
Vba = Vb+Va(Since they are of different dir.).
= 8+6.
= 14kmph.
Vb = 8kmph.
Vba = Vb+Va(Since they are of different dir.).
= 8+6.
= 14kmph.
(2)
Prem k said:
4 years ago
Efx = 6.
Efy = 8.
Resultant R= {sqr(Efx)+sqr(Efy)}1/2.
R = (36+64)'1/2.
R = (100)'1/2 = 10.
Efy = 8.
Resultant R= {sqr(Efx)+sqr(Efy)}1/2.
R = (36+64)'1/2.
R = (100)'1/2 = 10.
(1)
Alok kumar singh said:
4 years ago
I am not getting this, please Exlain in deeply.
Prashant said:
5 years ago
But how we can say 10km/h is the velocity relative to A?
Mohbat Baloch said:
6 years ago
By using Pythagoras theorem it is 10.
Amandeep yadav said:
6 years ago
6+(8/2)=6+4 = 10km/hr.
Kirthi said:
6 years ago
I am not getting How it's 10? please explain.
Biki said:
6 years ago
The Answer is correct. Then velocity of A relative to B is what? Is it also 10km/h?
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