A particle with a mass of `1 kg` a velocity of is having `10 m//s` in `+ve x`-direction at `t = 0`. Forces `vecF_(1)` and `vecF_(2)`, act on the particle whose magnitudes are changing with time according to the variation shown in Fig. The magnitude of the velocity of the particle at `t = 3 s` (neglect gravity effect) is found to be `root(n)(5)` Find the value of `n`

Answer

2

Explanation

`F_(1)=4N=-4hati 0lttle1s`
`=2N=-2hati, 1letle3ss`
` F_(2)1N=-hatj, 0lttle2s`
`=2t-3N-(2t-3)hatj 2letle3s`
Initial velocity of the particle is `vecu=10hati`
From impulse momentum theorem, `int vec(dp)=int vecF dt`
`vec("mv")-vec("mu")=int^(3)(vecF_(1)+vecF_(2))dt`
(where `vecv` is the required velocity)
`1xvecv-1xx10hati=-8hati-4hatj` or `vecv=2hati-4hatj`
or `v=sqrt(2^(2)+4^(2))=2sqrt(5) m//s`