Question

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

Answer 1

Hint: Firstly, you could do a vector representation of the data. Doing so, you get a clear picture of the motion here. Now you could recall the expression of relative velocity of one body with respect to the other. Now you could apply the formulae for vector subtraction to find the answer.

Formula used:
Expression for relative velocity of body A with respect to body B,
${{V}_{AB}}={{V}_{A}}-{{V}_{B}}$
Vector subtraction,
$\left|\overrightarrow{{{V}_{AB}}}\right|=\sqrt{{{V}_{A}}^{2}+{{V}_{B}}^{2}-2{{V}_{A}}{{V}_{B}}\cos \phi }$

Complete step-by-step answer:
In the question we are given two cars A and B. The car A is said to be moving towards east direction with a velocity of 10m/s and the car B is said to be moving toward north direction with a velocity of 20m/s. We are supposed to find the velocity of A with respect to B to an approximate value.
The concept that we are dealing here with is the relative velocity of two bodies in two dimensions. Let us see what relative velocity is of one body with respect to another in a general sense.
Let us consider two objects A and B, and they are moving with velocities ${{V}_{A}}$ and ${{V}_{B}}$ respectively (with respect to a common frame of reference, say ground). Then, the velocity of object A relative to that of object B is given by,
${{V}_{AB}}={{V}_{A}}-{{V}_{B}}$
Similarly, the velocity of object B relative to object A is given by,
${{V}_{BA}}={{V}_{B}}-{{V}_{A}}$
Clearly, ${{V}_{AB}}={{V}_{BA}}$
But, $\left| {{V}_{AB}} \right|=\left| {{V}_{BA}} \right|$
Now let us find the velocity of car A relative to the velocity of car B.

Relative velocity of car A with respect to B is given by,
$\overrightarrow{{{V}_{AB}}}=\overrightarrow{{{V}_{A}}}-\overrightarrow{{{V}_{B}}}$
And the magnitude of resultant vector is given by,
$\left| \overrightarrow{{{V}_{AB}}} \right|=\sqrt{{{V}_{A}}^{2}+{{V}_{B}}^{2}-2{{V}_{A}}{{V}_{B}}\cos \phi }$
Where $\phi $ is the angle between the vectors${{V}_{A}}$ and${{V}_{B}}$ , but they are perpendicular to each other,
$\phi =0$
$\Rightarrow \left| \overrightarrow{{{V}_{AB}}} \right|=\sqrt{{{V}_{A}}^{2}+{{V}_{B}}^{2}}$
In the given question,
${{V}_{A}}=10m/s$
${{V}_{B}}=20m/s$
$\Rightarrow \left| {{V}_{AB}} \right|=\sqrt{{{10}^{2}}+{{20}^{2}}}=\sqrt{500}$
$\Rightarrow \left| {{V}_{AB}} \right|=22.36m/s$
Therefore, the velocity of car A relative to car B is nearly 22m/s.

So, the correct answer is “Option C”.

Note: While dealing with problems related to motion in two dimensions, firstly you should make a vector representation of the given data in order to get a clear idea about the motion taking place. In the figure above we see that the resultant vector makes an angle θ with the horizontal. It can be given by,
$\tan \theta =\dfrac{{{V}_{B}}}{{{V}_{A}}}$
$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{{{V}_{B}}}{{{V}_{A}}} \right)={{\tan }^{-1}}\left( \dfrac{20}{10} \right)=63.43{}^\circ $