At time t=0, a particle is at (2 m, 4 m). It starts moving towards positive x-axis with constant acceleration 2 m/s2 (initial velocity = 0 ). After 2 s, an additional acceleration of 4 m/s2 starts acting on the particle in negative y-direction also. Find after next 2 s. coordinates of particle.
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a
x=16 m and y=-5 m
b
x=28 m and y=-4 m
c
x=25 m and y=-4 m
d
x=18 m and y=-4 m
answer is D.
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Detailed Solution
After 2 sv1=u+a1t1 =0+(2i)(2)=(4i^)m/s r1=ri+12a1t12 =(2i^+4j^)+12(2i^)(2)2 =(6i^+4j^)m(b)r2=r1+v1t2+12a2t22 =(6i^+4j^)+(4i^)(2)+12(2i^-4j^)(2)2 =(18i^-4j^)m∴ Co-ordinates are, x=18 m and y=-4 m