In going from one city to another, a car travels $75 k m$ north, $60 k m$ north-west and $20 k m$ east. The magnitude of displacement between the two cities is (take $\sqrt{2} = 0.7$ )

170 k m

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137 k m

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119 k m

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140 k m

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Solution

The correct option is C $119 k m$
Displacement of car $= 75^j$ ( $75$ km north)
$- 60 c o s 45^i + 60 s i n 45 j$ ( $60$ km north-west)
$+ 20 ↑$ ( $20$ km east)
$∴$ Final location $= (\frac{- 60}{\sqrt{2}} + 20) ↑ + (\frac{60}{\sqrt{2}} + 75)^j$
$= (- 60 \times 0.7 + 20) ↑ + (42 + 75)^j$
$= - 22 ↑ + 117^j$
displacement $= \sqrt{{(117)}^{2} + {(22)}^{2}} = \sqrt{14173} = 119$ km.

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In going from one city to another, a car travels 75km north, 60km north-west and 20km east. The magnitude of displacement between the two cities is (take √2=0.7)

The correct option is C 119kmDisplacement of car =75^j (75 km north) −60cos45^i+60sin45j (60 km north-west) +20↑ (20 km east)∴ Final location =(−60√2+20)↑+(60√2+75)^j =(−60×0.7+20)↑+(42+75)^j =−22↑+117^jdisplacement =√(117)2+(22)2=√14173=119 km.

In going from one city to another, a car travels $75 k m$ north, $60 k m$ north-west and $20 k m$ east. The magnitude of displacement between the two cities is (take $\sqrt{2} = 0.7$ )