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A particle of mass 2 kg moves in the xy ...

A particle of mass `2 kg` moves in the `xy` plane under the action of a constant force `vec(F)` where `vec(F)=hat(i)-hat(j)`. Initially the velocity of the particle is `2hat(j)`. The velocity of the particle at time `t` is

A

`(1)/(2)(t+4)hat(i)-(1)/(2)that(j)`

B

`t(hat(i)-hat(j))`

C

`(1)/(2)t(hat(i)-hat(j))`

D

`(1)/(2)that(i)+(1)/(2)(t+4)hat(j)`

Text Solution

Verified by Experts

The correct Answer is:
A
`m=2kg,vec(F)=hat(i)-hat(j)`.
`rArr vec(a)=(vec(F))/(m)=(1)/(2)(hat(i)-hat(j))`
Now `vec(V)=vec(u)+vec(a)t.`
`rArr vec(V)=2hat(i)+(1)/(2)(hat(i)-hat(j))t.`
`=(2+(t)/(2))hat(i)-(t)/(2)hat(j)=(1)/(2)(t+4)hat(i)-(t)/(2)hat(j)`
Alter `:` Substitute `t=0` in option and get answer
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Knowledge Check

  • An object of mass 2 kg at rest at origin starts moving under the action of a force vec(F)=(3t^(2)hat(i)+4that(j))N The velocity of the object at t = 2 s will be -

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    C
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    D
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