At time `t= 0,` a particle is at `(2m, 4m).` It starts moving towards positive x-axis with constant acceleration `2 m//s^2` (initial velocity=0). After 2 s, an additional acceleration of `4 m//s^2` starts acting on the particle in negative y-direction also. Find after next 2 s.
(a) velocity and
(b) coordinates of particle.

Answer

A, B, D

Explanation

After 2s
`v_1=u+a_1t_1`
`=0+(2 hati)(2)=(4 hati) m//s`
`r_1=1_i+1/2a_1t_1^2`
`=(2 hati+4 hatj)+1/2(2hati)(2)^2`
`=(6 hati+4 hatj)m`
After next 2s
(a) `v_2=v_1+a_2t_2`
`=(4 hati)+(2hati-4hatj)(2)`
`=(8hati-8 hatj)m//s`
(b)`r_2=r_1+v_1t_2+1/2a_2t_2^2`
`=(6 hati+4 hatj)+(4 hati)(2)+1/2(2 hati-4 hatj)(2)^2`
`=(18 hati-4 hatj)m `
`:. ` Co-ordinates are,
`x=18 m`
and `y=-4 m`