A particle is initially at 20 - m- Now particle starts to move with velocity 2 -2 -3- m - s- Find distance of particle from origin after 5 seconds-A- 30 mB- 40 -3 m- 34 mD- 28 m

The correct option is C 34 mr⃗ = nbsp;A⃗ + nbsp;B⃗ r⃗ = 20 î + 10 î + 10 √(3)ĵ = 30 î + 10 √(3) nbsp;î |r⃗| = √(900 + 300) = √(1200) = 10 √(12) = 20 √(3) ...

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Standard XII

Physics

Angular Momentum

A particle is...

Question

A particle is initially at $20^i$ m. Now particle starts to move with velocity $2^i + 2 \sqrt{3}^j$ m/s. Find distance of particle from origin after 5 seconds.

30

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40 \sqrt{3}

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34

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28

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Solution

The correct option is C $34$ m
$\to r = \to A + \to B$
$\to r = 20^i + 10^i + 10 \sqrt{3}^j$
$= 30^i + 10 \sqrt{3}^i$
$| \to r | = \sqrt{900 + 300} = \sqrt{1200} = 10 \sqrt{12}$
$= 20 \sqrt{3}$
$= 34$ m

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A particle is initially at 20^im. Now particle starts to move with velocity 2^i+2√3^j m/s. Find distance of particle from origin after 5 seconds.

The correct option is C 34 m→r=→A+→B →r=20^i+10^i+10√3^j =30^i+10√3^i |→r|=√900+300=√1200=10√12 =20√3 =34 m

A particle is initially at $20^i$ m. Now particle starts to move with velocity $2^i + 2 \sqrt{3}^j$ m/s. Find distance of particle from origin after 5 seconds.

The correct option is C $34$ m
$\to r = \to A + \to B$
$\to r = 20^i + 10^i + 10 \sqrt{3}^j$
$= 30^i + 10 \sqrt{3}^i$
$| \to r | = \sqrt{900 + 300} = \sqrt{1200} = 10 \sqrt{12}$
$= 20 \sqrt{3}$
$= 34$ m