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Q.

A spy report about a suspected car reads as follows. “The car moved 2km towards east, made a perpendicular left turn, ran for 500m, made a perpendicular right turn, ran for 4km and stopped. If the magnitude and direction of the displacement of the car is $\frac{12.04}{n} and {Tan}^{- 1} (\frac{1}{6 n})$ find the value of n

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answer is 2.

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Detailed Solution

$\vec{OC} = \vec{OA} + \vec{AB} + \vec{BC} = 2 \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{i} = 6 \hat{i} + \frac{1}{2} \hat{j}$

$| \vec{O C} | = \sqrt{36 + \frac{1}{4}} = \frac{\sqrt{145}}{2} = \frac{12.04}{2}$

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$Tan θ = (\frac{1 / 2}{6}) = \frac{1}{12}; θ = {Tan}^{- 1} (\frac{1}{12}), ∴ n = 2$

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A spy report about a suspected car reads as follows. “The car moved 2km towards east, made a perpendicular left turn, ran for 500m, made a perpendicular right turn, ran for 4km and stopped. If the magnitude and direction of the displacement of the car is 12.04n and Tan−1⁡16n find the value of n