Answer
406.2k+ views
Hint
Here, we have to find out the velocity of the velocity of A w.r.t B i.e. $V_AB$ which is equal to the difference of final velocity and initial velocity i.e. $\mathop {{V_{AB}}}\limits^ \to = \mathop {{V_B}}\limits^ \to - \mathop {{V_A}}\limits^ \to $,where $\mathop {{V_A}}\limits^ \to = 10\hat i$and $\mathop {{V_B}}\limits^ \to = 20\hat j$, for finding the velocity we have to find out the magnitude of the velocity which can be determined as $\left| {\mathop {{V_{AB}}}\limits^ \to } \right| = \sqrt {V_A^2 + V_B^2} $, on substituting the values we get the desired result.
Complete step by step answer
it is given that, car A is moving towards east with velocity 10 m/s,
car B is moving towards the north with velocity 20 m/s.
we have to find out the value of the velocity of car A with respect to B i.e. $V_AB$
let the unit vector of east direction is so the velocity of car A is $\mathop {{V_A}}\limits^ \to = 10\hat i$
similarly, the unit vector of north direction is , the velocity of car B is $\mathop {{V_B}}\limits^ \to = 20\hat j$
now, the velocity of car A with respect to car B is $\mathop {{V_{AB}}}\limits^ \to $, which is equal to the difference between the final and initial velocity i.e. $\mathop {{V_{AB}}}\limits^ \to = \mathop {{V_B}}\limits^ \to - \mathop {{V_A}}\limits^ \to $
substitute the values, we get
$ \Rightarrow \mathop {{V_{AB}}}\limits^ \to = 10\hat i - 20\hat j$ …………………. (1)
The magnitude of the velocity of car A with respect to car B is
$\left| {\mathop {{V_{AB}}}\limits^ \to } \right| = \sqrt {V_A^2 + V_B^2} $
Using equation (1), we get
$ \Rightarrow \left| {\mathop {{V_{AB}}}\limits^ \to } \right| = \sqrt {{{\left( {10} \right)}^2} + {{\left( { - 20} \right)}^2}} = \sqrt {500} $
$ \Rightarrow \left| {\mathop {{V_{AB}}}\limits^ \to } \right| = 10\sqrt 5 $
$ \Rightarrow \left| {\mathop {{V_{AB}}}\limits^ \to } \right| \approx 22m{s^{ - 1}}$
This is the desired velocity of car A with respect to car B.
Hence, option (C) is correct.
Note
The relative velocity of an object A with respect to another object B is the velocity that object A would appear to have to an observer situated on object B moving along with it or vice versa.
The magnitude of the relative velocity of two objects making an angle θ is $\left| {\mathop {{V_{AB}}}\limits^ \to } \right| = \sqrt {V_A^2 + V_B^2 - 2{V_A}{V_B}\cos \theta } $ , in above case as north and east direction are at 900 angle and therefore formula will become, $\left| {\mathop {{V_{AB}}}\limits^ \to } \right| = \sqrt {V_A^2 + V_B^2} $
Here, we have to find out the velocity of the velocity of A w.r.t B i.e. $V_AB$ which is equal to the difference of final velocity and initial velocity i.e. $\mathop {{V_{AB}}}\limits^ \to = \mathop {{V_B}}\limits^ \to - \mathop {{V_A}}\limits^ \to $,where $\mathop {{V_A}}\limits^ \to = 10\hat i$and $\mathop {{V_B}}\limits^ \to = 20\hat j$, for finding the velocity we have to find out the magnitude of the velocity which can be determined as $\left| {\mathop {{V_{AB}}}\limits^ \to } \right| = \sqrt {V_A^2 + V_B^2} $, on substituting the values we get the desired result.
Complete step by step answer
it is given that, car A is moving towards east with velocity 10 m/s,
car B is moving towards the north with velocity 20 m/s.
we have to find out the value of the velocity of car A with respect to B i.e. $V_AB$
let the unit vector of east direction is so the velocity of car A is $\mathop {{V_A}}\limits^ \to = 10\hat i$
similarly, the unit vector of north direction is , the velocity of car B is $\mathop {{V_B}}\limits^ \to = 20\hat j$
now, the velocity of car A with respect to car B is $\mathop {{V_{AB}}}\limits^ \to $, which is equal to the difference between the final and initial velocity i.e. $\mathop {{V_{AB}}}\limits^ \to = \mathop {{V_B}}\limits^ \to - \mathop {{V_A}}\limits^ \to $
substitute the values, we get
$ \Rightarrow \mathop {{V_{AB}}}\limits^ \to = 10\hat i - 20\hat j$ …………………. (1)
The magnitude of the velocity of car A with respect to car B is
$\left| {\mathop {{V_{AB}}}\limits^ \to } \right| = \sqrt {V_A^2 + V_B^2} $
Using equation (1), we get
$ \Rightarrow \left| {\mathop {{V_{AB}}}\limits^ \to } \right| = \sqrt {{{\left( {10} \right)}^2} + {{\left( { - 20} \right)}^2}} = \sqrt {500} $
$ \Rightarrow \left| {\mathop {{V_{AB}}}\limits^ \to } \right| = 10\sqrt 5 $
$ \Rightarrow \left| {\mathop {{V_{AB}}}\limits^ \to } \right| \approx 22m{s^{ - 1}}$
This is the desired velocity of car A with respect to car B.
Hence, option (C) is correct.
Note
The relative velocity of an object A with respect to another object B is the velocity that object A would appear to have to an observer situated on object B moving along with it or vice versa.
The magnitude of the relative velocity of two objects making an angle θ is $\left| {\mathop {{V_{AB}}}\limits^ \to } \right| = \sqrt {V_A^2 + V_B^2 - 2{V_A}{V_B}\cos \theta } $ , in above case as north and east direction are at 900 angle and therefore formula will become, $\left| {\mathop {{V_{AB}}}\limits^ \to } \right| = \sqrt {V_A^2 + V_B^2} $
Recently Updated Pages
In a flask the weight ratio of CH4g and SO2g at 298 class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In a flask colourless N2O4 is in equilibrium with brown class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In a first order reaction the concentration of the class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In a first order reaction the concentration of the class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In a fermentation tank molasses solution is mixed with class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In a face centred cubic unit cell what is the volume class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name 10 Living and Non living things class 9 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
List some examples of Rabi and Kharif crops class 8 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)