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Question

# Match Column - I with Column -II Column - IColumn - IIi.Velocity of a particle following (p) αequation x=u(t−8)+α(t−2)2at t=0ii.Acceleration of a particle moving according to equation(q)2αx=u(t−8)+α(t−2)2 at t=0iii.Velocity of the particle moving (r) uaccording to equation x=ut+5αt2ln(1+t2)+αt2at t=0iv.Acceleration of the particle moving (s) u−4αaccording to the equation x=ut+5αt2(1+t2)+αt2at t=0

A
i(s);ii(q);iii(r);iv(q)
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B
i(s);ii(q);iii(s);iv(p)
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C
i(r);ii(p);iii(s);iv(q)
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D
i(r);ii(r);iii(r);iv(q)
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Solution

## The correct option is A i→(s);ii→(q);iii→(r);iv→(q)i. x=u(t−8)+α(t−2)2 Differentiating w.r.t t, we get v=u+2α(t−2)×1 ⇒v=u+2α(t−2) At t=0 v=u−4α ii. x=u(t−8)+α(t−2)2 Differentiating w.r.t t we get v=u+2α(t−2) Again differentiating w.r.t t we get a=2α At t=0, a=2α iii. x=ut+5αt2ln(1+t2)+αt2 Differentiating w.r.t t , we get v=u+10 αtln(1+t2)+5.αt2.121+t2+2αt =u+10αtln(1+t2)+5αt22+t+2αt At t=0, v=u+0+0+0=u iv. x=ut+5αt2ln(1+t2)+αt2 ⇒v=u+10 αtln(1+t2)+5αt22+t+2at Differentiating w.r.t t, we get a=10α.ln(1+t2)+10αt.121+t2+10αt(2+t)−5αt2(2+t)2+2α At t=0 a=0+0+0+2α=2α

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