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:
4 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.
(6)
ABDUL SALAM ANSARI said:
1 decade ago
Both force are acting on right angle so, we will find resultant according to parallelogram law.
R^2 = f1^2 + f2^2 +2f1*f2*cos90.
But there is value of cos90 = 0.
We know,
f1 = 6km/hr and f2 = 8km/hr.
R^2 = 6^2+8^2+2*6*8*0.
=> R^2 = 36+64.
=> R = (100)^1/2.
=> R = 10km/hr.
R^2 = f1^2 + f2^2 +2f1*f2*cos90.
But there is value of cos90 = 0.
We know,
f1 = 6km/hr and f2 = 8km/hr.
R^2 = 6^2+8^2+2*6*8*0.
=> R^2 = 36+64.
=> R = (100)^1/2.
=> R = 10km/hr.
CHANDAN KUMAR said:
1 decade ago
The force are acting at right angle so simply you know that find resultant r^2 = p^2+Q^2.
Assume p = 6, q = 8 substitute this values.
r^2 = 6^2+8^2.
r^2 = 36+64.
r^2 = 100.
Now do the square root of 100 is 10.
Thus r = 10km/hr.
Assume p = 6, q = 8 substitute this values.
r^2 = 6^2+8^2.
r^2 = 36+64.
r^2 = 100.
Now do the square root of 100 is 10.
Thus r = 10km/hr.
Saleem khan said:
1 decade ago
Both force are acting on right angle so we will find resultant acc to parallelogram law. R^2 = f1^2 + f2^2 + cos90.
But there is value of cos90 = 0.
But there is value of cos90 = 0.
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.
A KUMAR said:
1 decade 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.
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)
Gaurav gupta said:
1 decade ago
By using pythagoras theorem;
(8*8) + (6*6) equal to 100 and square root of 100 is 10.
(8*8) + (6*6) equal to 100 and square root of 100 is 10.
Biki said:
8 years ago
The Answer is correct. Then velocity of A relative to B is what? Is it also 10km/h?
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