Question

Two cars are moving, car $A$ along east and car $B$ is moving along south. Draw the direction of the motion of
(a) car $A$ as seen from car $B$
(b) car $B$ as seen from car $A$

Accepted Answer

$(a)$ Considering direction of velocity of car $A$ as positive.

The relative velocity of $A$ with respect to $B$ is ${\vec{v}}_{AB} = {\vec{v}}_{A} - {\vec{v}}_{B} = {\vec{v}}_{A} + (- {\vec{v}}_{B})$

Magnitude of relative velocity is $|{\vec{v}}_{A, B}| = \sqrt{v_{A}^{2} + v_{B}^{2}}$

If $θ$ be the angle between them, then direction of motion is $\tan θ = \frac{v_{B}}{v_{A}}$

$θ = \tan^{- 1} (\frac{v_{B}}{v_{A}})$ .

The direction of motion of car $A$ as seen from car $B$ is shown below.

$(b)$ Considering direction of velocity of car $B$ as positive.

The relative velocity of $B$ with respect to $A$ is ${\vec{v}}_{BA} = {\vec{v}}_{B} - {\vec{v}}_{A} = {\vec{v}}_{B} + (- {\vec{v}}_{A})$

Magnitude of relative velocity is $|{\vec{v}}_{B, A}| = \sqrt{v_{A}^{2} + v_{B}^{2}}$

Direction of motion is $\tan θ = \frac{v_{A}}{v_{B}} \Rightarrow θ = \tan^{- 1} (\frac{v_{A}}{v_{B}})$

The direction of motion of car $B$ as seen from car $A$ is shown below.

B M Sharma Solutions for Chapter: Kinematics II, Exercise 45: ILLUSTRATION 5.41

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics II, Exercise 45: ILLUSTRATION 5.41

Questions from B M Sharma Solutions for Chapter: Kinematics II, Exercise 45: ILLUSTRATION 5.41 with Hints & Solutions

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