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A 0.2 kg object at rest is subjected to ...

A `0.2 kg` object at rest is subjected to a force `(0.3hati-0.4hatj) N`. What is its velocity vector after `6 sec`

A

`( 9 hati -12 hatj)`

B

`( 8 hati - 16 hatj)`

C

`(12 hati - 9 hatj)`

D

`(16 hati - 8 hatj)`

Text Solution

Verified by Experts

The correct Answer is:
A
Here ` m= 0.2 kg ,u= I , vecF = ( 0.3 hati - 0.4 hatj ) N ,i=6 s `
` therefore veca = (vecF)/(m ) =((0.3 hati - 0.4 hatj ))/( 0.2 ) =(3)/(2) hati - 2 hatj`
` vecv= vecu + veca t =0 +((3)/(2) hati - 2 hatj) xx 6 = 9 hati - 12 hatj`
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