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The displacement of a particle moving in a straight line is described by the relation S=6+12t-2t2. Here s in meter and t is in sec. The distance covered by the particle in first 5s is

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detailed solution

Correct option is D

s=6+12t−2t2differentiating with respect to tv=12−4tdirection is reversed , v=0at time, t = 3sAt t=0,The displacement is 6 metersdisplacement in 3 secs is 6+12(3)−2(3)2=24 displacement in 5 secs is 6+12(5)−2(5)2=16 Distance travelled in forward direction in 3 sec is 24-6=18 mDistance travelled in reverse direction in next 2 sec is 24-16=8 mThus total distance travelled in 5 sec= 18+8=26 m

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