First slide

Projection Under uniform Acceleration

Question

As two boats approach the Mumbai, the velocity of boat I relative to boat 2 is 10 $\sqrt{3}$ kmhr^-1 in a direction of 60o north of east. If boat 2 has a velocity of l5 kmhr^-1 due south. What is the velocity of boat 1 ?

Easy

A

5 $\sqrt{3}$ krnhr^-1 due south-east
B

5 knhr^-1 due east
C

l0 kmhr^-1 due north
D

5 $\sqrt{3}$ kmhr^-1 due east

Solution

${\vec{v}}_{12} = {\vec{v}}_{1} - {\vec{v}}_{2} \Rightarrow {\vec{v}}_{1} = {\vec{v}}_{2} + {\vec{v}}_{12}$
Here, ${\vec{v}}_{12} = 10 \sqrt{3} (\cos 60^{\circ} \hat{i} + \sin 60^{\circ} \hat{j}) = 5 \sqrt{3} \hat{i} + 15 \hat{j}$
$\begin{array}{l} \Rightarrow {\vec{v}}_{1} = {\vec{v}}_{2} + {\vec{v}}_{12} = (- 15 \hat{j}) + (5 \sqrt{3} \hat{i} + 15 \hat{j}) = 5 \sqrt{3} \hat{i} \\ \Rightarrow {\vec{v}}_{1} = 5 \sqrt{3} km / hr due east . \end{array}$

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