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Byju's Answer

Standard XII

Physics

Applications of Relative Projectile

A particle A ...

Question

A particle (A) moves due north at 3 km/h, another particle (B) due west at 4 km/h. The relative velocity of A with respect to B is $(t a n 37^{\circ} = 3 / 4)$

5 k m / h, 37^{\circ}

north of east

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5 k m / h, 37^{\circ}

east of north

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5 \sqrt{2} k m / h, 53^{\circ}

east of north

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5 \sqrt{2} k m / h, 53^{\circ}

north of east

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Solution

The correct option is A $5 k m / h, 37^{\circ}$ north of east
$_{A B} =_{A} -_{B}$

$V_{A B} = \sqrt{V_{A}^{2} + V_{B}^{2}} = 5 k m / h$
$α = t a n^{- 1} (\frac{V_{A}}{V_{B}})$
$= t a n^{- 1} (\frac{3}{4}) = 37^{\circ}$ .

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A particle (A) moves due north at 3 km/h, another particle (B) due west at 4 km/h. The relative velocity of A with respect to B is (tan 37∘=3/4)

The correct option is A 5 km/h,37∘north of east−−→VAB=−→VA−−→VB VAB=√V2A+V2B=5 km/h α=tan−1(VAVB) =tan−1(34)=37∘.

A particle (A) moves due north at 3 km/h, another particle (B) due west at 4 km/h. The relative velocity of A with respect to B is $(t a n 37^{\circ} = 3 / 4)$

The correct option is A $5 k m / h, 37^{\circ}$ north of east
$_{A B} =_{A} -_{B}$

$V_{A B} = \sqrt{V_{A}^{2} + V_{B}^{2}} = 5 k m / h$
$α = t a n^{- 1} (\frac{V_{A}}{V_{B}})$
$= t a n^{- 1} (\frac{3}{4}) = 37^{\circ}$ .