Question

A car travels $6 km$ towards north at an angle of $45 °$ to the east and then travels distance of $4 km$ towards north at an angle of $135 °$ to the east. How far is the point from the starting point? What angle does the straight line joining its initial and final position makes with the east?

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Accepted Answer

Here car travels in two steps:
In the first step, it travels $6 km$ in the north making an angle of $45 °$ to the east.
Taking components for this step,
The magnitude of the displacement is $135 °$
So, component in $X$ direction is $d_{1 x} = d \cos (45) = \frac{6}{\sqrt{2}}$
The component in the $Y$ direction is $d_{1 y} = d \sin (45 °) = \frac{6}{\sqrt{2}}$
Hence, the displacement vector, $\vec{d_{1}} = d_{1 x} \hat{i} + d_{2 y} \hat{j} = \frac{6}{\sqrt{2}} \hat{i} + \frac{6}{\sqrt{2}} \hat{j}$
Similarly, for the second step, $d_{2} = 4 km$ and angle with the east is $135 °$ .
So, $d_{2 x} = d_{2} \cos (135 °) = \frac{- 4}{\sqrt{2}}$ and $d_{2 y} = d_{2} \sin (135 °) = \frac{4}{\sqrt{2}}$
Hence, displacement vector for the second step, $\vec{d_{2}} = d_{2 x} \hat{i} + d_{2 y} \hat{j} = \frac{- 4}{\sqrt{2}} \hat{i} + \frac{4}{\sqrt{2}} \hat{j}$

Now, the net displacement,

$\vec{d} = \vec{d_{1}} + {\vec{d}}_{2}$
$\vec{d} = (\frac{6}{\sqrt{2}} \hat{i} + \frac{6}{\sqrt{2}} \hat{j}) + (\frac{- 4}{\sqrt{2}} \hat{i} + \frac{4}{\sqrt{2}} \hat{j})$
$\Rightarrow \vec{d} = \frac{2}{\sqrt{2}} \hat{i} + \frac{10}{\sqrt{2}} \hat{j}$
The magnitude of net displacement, $d = \sqrt{{(\frac{2}{\sqrt{2}})}^{2} + {(\frac{10}{\sqrt{2}})}^{2}} = \sqrt{2 + 50} = \sqrt{52} km$
The angle made by the net displacement vector with the east is, $\tan (θ) = \frac{y}{x} = \frac{\frac{10}{\sqrt{2}}}{\frac{2}{\sqrt{2}}} = 5$
$\Rightarrow θ = \tan^{- 1} (5)$

B M Sharma Solutions for Chapter: Vectors, Exercise 34: CONCEPT APPLICATION EXERCISE 3.3

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Vectors, Exercise 34: CONCEPT APPLICATION EXERCISE 3.3

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