Answer the question briefly

Dear Student,

Please find below the solution to the asked query:

Given that the acceleration of the particle changes with time as,

So, velocity varies with time as,

V=V0+a dt = 3t2+2t+2dtV=V0+t3+t2+2tat t=0; V=2 msV=t3+t2+2t+2

So, at t = 2 sec.

Vt=t3+t2+2t+2  Vt=2=23+22+2×2+2=8+4+4+2 Vt=2=18 ms

Position of particle at any time is,

X=X0+Vt dt = 2+02t3+t2+2t+2dt X=2+t44+t33+t2+2t02=2+164+83+4+4=2+4+8+83 X=503 m


Hope this information will clear your doubts about the topic.

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