Question

Out of two cars A and B, car A is moving towards east with a velocity of 10 m/s whereas B is moving towards north with a velocity 20 m/s, then velocity of A w.r.t B is (nearly)
(A) 30m/s
(B) 10m/s
(C) 22m/s
(D) 42m/s

Answer 1

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, $\left| {\mathop {{V_{AB}}}\limits^ \to } \right| = \sqrt {V_A^2 + V_B^2}$