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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Infinity Learn · Accepted Answer

$$s=6+12t$$−$$2t2differentiating$$ 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

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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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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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