### Discussion :: Applied Mechanics - Section 1 (Q.No.3)

Gaurav Gupta said: (Apr 26, 2013) | |

By using pythagoras theorem; (8*8) + (6*6) equal to 100 and square root of 100 is 10. |

Saurabh said: (Jul 22, 2013) | |

It should be 10. Why it is 14 anyone explain this? |

Ankur Padalia said: (Dec 2, 2013) | |

Using the pythagoras theorem. |

Amol Ashok Lande said: (Feb 6, 2014) | |

(V)x = 6. (V)y = 8. Vb/a = (36+64)^1/2. Vb/a = 10 km/hr. |

Saleem Khan said: (Apr 21, 2014) | |

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. |

Abdul Salam Ansari said: (Feb 6, 2015) | |

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. |

Chandan Kumar said: (Feb 16, 2015) | |

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. |

A Kumar said: (Jul 25, 2015) | |

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. |

Usama Ahmad said: (Aug 19, 2015) | |

6*6+8*8=100 s of 100=10. |

Nitesh Jayswal said: (Dec 15, 2016) | |

By using Pythagoras theorem:- 8 * 8 + 6 * 6 = 100. Square root of 100 = 10km/hr. |

Kuldeep Singh Umath said: (Jan 8, 2017) | |

It should be 10. I agree. |

Jitendra Singh said: (Jan 14, 2017) | |

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. |

Biki said: (Jun 3, 2017) | |

The Answer is correct. Then velocity of A relative to B is what? Is it also 10km/h? |

Kirthi said: (Jul 16, 2017) | |

I am not getting How it's 10? please explain. |

Amandeep Yadav said: (Sep 8, 2017) | |

6+(8/2)=6+4 = 10km/hr. |

Mohbat Baloch said: (Sep 10, 2017) | |

By using Pythagoras theorem it is 10. |

Prashant said: (Mar 12, 2018) | |

But how we can say 10km/h is the velocity relative to A? |

Alok Kumar Singh said: (Feb 19, 2019) | |

I am not getting this, please Exlain in deeply. |

Prem K said: (Jun 4, 2019) | |

Efx = 6. Efy = 8. Resultant R= {sqr(Efx)+sqr(Efy)}1/2. R = (36+64)'1/2. R = (100)'1/2 = 10. |

Larry L said: (Nov 6, 2019) | |

Va = 6kmph. Vb = 8kmph. Vba = Vb+Va(Since they are of different dir.). = 8+6. = 14kmph. |

Onkar said: (Oct 1, 2020) | |

Va = 6. Vb = 8. Va/b = √(va+vb)= √6^2 + 8^2 = √100. Va/b = 10. |

Nabin Acharya said: (Jun 27, 2021) | |

@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. |

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