The force on a particle of mass `10g` is `(hati 10+hatj 5)`N If it starts from rest what would be its position at time `t=5s` ?

Options

A. `12500 hati +6250 hatj m`
B. `6250 hati + 12500 hatj m`
C. `12500 hati + 12500 hatj m`
D. `6250 hati + 6250 hatj m`

Answer

A

Explanation

From `vecF = (10 hati + 5 hatj ) N` we have
` F_x = 10 N, F_y = 5N `
` therefore a_x = (F_x)/( m ) = ( 10)/( 0.01 ) = 1000 ms^(-2)`
As acceleration along x-axis is constant ,
` therefore x= u_x t + 1/2 a_y t^2 = 0 + 1/2 xx 500 xx 5^2 = 6250 m`
hence , position of particle at t=5 s is
` vecr = (12500 hati + 6250 hatj ) m`