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 2 of 3.
Kuldeep Singh umath said:
9 years ago
It should be 10. I agree.
Jitendra Singh said:
9 years ago
Va = 6i+0j, Vb = 0i + 8j.
Vba = Vb - Va.
6i + 0j - (0i + 8j) = 6i - 8j.
Magnitude of Vba = (6^2+8^2)^1/2.
= (100)^1/2.
= 10 mtr/sec.
Vba = Vb - Va.
6i + 0j - (0i + 8j) = 6i - 8j.
Magnitude of Vba = (6^2+8^2)^1/2.
= (100)^1/2.
= 10 mtr/sec.
Biki said:
8 years ago
The Answer is correct. Then velocity of A relative to B is what? Is it also 10km/h?
Kirthi said:
8 years ago
I am not getting How it's 10? please explain.
Amandeep yadav said:
8 years ago
6+(8/2)=6+4 = 10km/hr.
Mohbat Baloch said:
8 years ago
By using Pythagoras theorem it is 10.
(1)
Prashant said:
7 years ago
But how we can say 10km/h is the velocity relative to A?
Alok kumar singh said:
7 years ago
I am not getting this, please Exlain in deeply.
Prem k said:
6 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)
Larry L said:
6 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)
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