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A car moves from O to D along the path
OABCD shown in fig.
What is distance travelled and its net displacement ?

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Distance $= 17 k m$ , Displacement $= 5 k m, 37 ° S o f E$

Distance $= 16 k m$ , Displacement $= 3 k m, 37 ° S o f E$

Distance $= 15 k m$ , Displacement $= 4 k m, 37 ° S o f E$

Distance $= 18 k m$ , Displacement $= 6 k m, 37 ° S o f E$

answer is C.

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Detailed Solution

Distance $= |\vec{O A}| + |\vec{A B}| + |\vec{B C}| + |\vec{C D}|$

$= 8 + 4 + 4 + 1 = 17 k m$

Displacement $= \vec{O A} + \vec{A B} + \vec{B C} + \vec{C D}$

$= 8 \hat{i} + (- 4 \hat{j}) + (- 4 \hat{i}) + \hat{j} = 4 \hat{i} - 3 \hat{j}$

$\Rightarrow |d i s p l a c e m e n t| = \sqrt{{(4)}^{2} + {(3)}^{2}} = 5$

So, Displacement $= 5 k m, 37 ° S o f E$

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