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.

At time `t= 0,` a particle is at `(2m, 4m).` It starts moving towards

1.	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» Correct Answer - A::B::D 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`

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