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Standard XII

Physics

Vector Component

If a particle...

Question

If a particle $A$ is moving along a straight line with velocity $3 m / s$ and another particle $B$ has a velocity $5 m / s$ at an angle of $60^{o}$ to the path of $A$ , then the velocity $B$ relative to $A$ is

\sqrt{39} m / s

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\sqrt{19} m / s

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19 m / s

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None of these

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Solution

The correct option is B $\sqrt{19} m / s$
Given, $V_{A} = 3 m / s$
and $V_{B} = 5 m / s$
$∴ V_{B A} = \sqrt{V_{A}^{2} + V_{B}^{2} + 2 V_{A} V_{B} cos θ}$ $(∵ θ = 120^{o})$
$= \sqrt{9 + 25 + 2 \times 3 \times 5 \times (- \frac{1}{2})}$
$= \sqrt{34 - 15} = \sqrt{19} m / s$ .

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If a particle A is moving along a straight line with velocity 3m/s and another particle B has a velocity 5m/s at an angle of 60o to the path of A, then the velocity B relative to A is

The correct option is B √19m/sGiven, VA=3m/sand VB=5m/s∴VBA=√V2A+V2B+2VAVBcosθ (∵θ=120o) =√9+25+2×3×5×(−12) =√34−15=√19m/s.

If a particle $A$ is moving along a straight line with velocity $3 m / s$ and another particle $B$ has a velocity $5 m / s$ at an angle of $60^{o}$ to the path of $A$ , then the velocity $B$ relative to $A$ is

The correct option is B $\sqrt{19} m / s$
Given, $V_{A} = 3 m / s$
and $V_{B} = 5 m / s$
$∴ V_{B A} = \sqrt{V_{A}^{2} + V_{B}^{2} + 2 V_{A} V_{B} cos θ}$ $(∵ θ = 120^{o})$
$= \sqrt{9 + 25 + 2 \times 3 \times 5 \times (- \frac{1}{2})}$
$= \sqrt{34 - 15} = \sqrt{19} m / s$ .