Question

The displacement of a particle moving in a straight line is described by the relation, s=6+12tâˆ’2t2. Here â€²sâ€² is in metre and â€²tâ€² is in second. The distance covered by particle in first 5 sec is:

A
20 m
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B
32 m
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C
24 m
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D
26 m
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Solution

The correct option is D 26 ms=6+12t−2t2differentiating with respect to tv=12−4tdirection is reversed v=0At t=0,The displacement is 6 metersdisplacement in 3 secs is 6+12times3−2times32=24displacement in 5 secs is 6+12times5−2times52=16Distance 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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