A particle moving with uniform retardation covers distances $$18m$$, $$14m$$ & $$10m$$ in successive seconds. It comes to rest after travelling a further distance of

Question

Infinity Learn · Accepted Answer

At time $$t=0$$     u is velocitygiven distance in successive seconds as $$18m$$,$$14m$$,$$10mlet$$ initial velocity $$=u$$; acceleration $$=a$$; distance covered in nth second $$isS=u+a22n$$−$$1put$$ $$n=1$$→$$S1=u+a2(2$$×$$1-1)18=u+a2--$$→(1)S2=u+a22n$$−$$1put$$ $$n=2$$→$$14=u+a22$$×$$2$$−$$114=u+32a--$$→$$(2)solving$$ equations $$(1)&$$(2)u=20m/sa=-4m/s2by$$ question,particle finally comes to rest $$v=0total$$ distance travelled by the body  $$s=v2-u22a$$ ⇒ $$s=02-2022(-4)s=50m=total$$ distance travelled by the particleIt stops after travelling further distance after the three successive seconds $$=50-(18+14+10)=8m$$

A particle moving with uniform retardation covers distances 18m, 14m & 10m in successive seconds. It comes to rest after travelling a further distance of

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