Question

A ship is moving at 40 km.h-1 towards west. Another
is proceeding southward at 30 km-h-1 . Find the rela-
tive velocity of the second ship with respect to the first.
[150 km.h-1, inclined towards east making an angle
tan-1 with the south]​

Answer 1

Given :

Velocity of ship B = 40km/h (West)

Velocity of ship A = 30km/h (South)

To Find :

Relative velocity of ship B wrt ship A.

Solution :

■ Relative velocity of object B wrt A is given by

• $\boxed{\tt{\orange{\overrightarrow{V_{BA}}=\overrightarrow{V_B}-\overrightarrow{V_A}}}}$

$:\implies\sf\:\overrightarrow{V_A}=40\:(\hat{-j})$

$:\implies\sf\:\overrightarrow{V_B}=30\:(\hat{-i})$

Calculation of relative velocity :

$:\implies\tt\:\overrightarrow{V_{BA}}=\overrightarrow{V_B}-\overrightarrow{V_A}$

$:\implies\tt\:\overrightarrow{V_{BA}}=\overrightarrow{V_B}+(-\overrightarrow{V_A})$

$\dag\tt\:\:-\overrightarrow{V_A}=40\:\hat{i}$

$:\implies\sf\:\overrightarrow{V_{BA}}=\sqrt{30^2+40^2}\:(\hat{-j}+\hat{i})$

$:\implies\sf\:\overrightarrow{V_{BA}}=\sqrt{900+1600}\:(\hat{-j}+\hat{i})$

$:\implies\sf\:\overrightarrow{V_{BA}}=\sqrt{2500}\:(\hat{-j}+\hat{i})$

$:\implies\bf\:\overrightarrow{V_{BA}}=50\:(\hat{-j}+\hat{i})$

Magnitude : $\underline{\boxed{\bf{\purple{|\overrightarrow{V_{BA}}|=50\:ms^{-1}}}}}$

■ Direction :

$:\implies\sf\:tan\theta=\dfrac{V_A}{V_B}$

$:\implies\sf\:tan\theta=\dfrac{40}{30}$

$:\implies\:\underline{\boxed{\bf{\pink{\theta=tan^{-1}\big(\dfrac{4}{3}\big)\:towards\:SE}}}}$

Answer 2

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