Acceleration of the particle in a rectilinear motion is given as a=4t+3 in (m/s2). The final position of the particle as a function of time t is (Given, at t=0,u=3m/s,x=2m)
A
2t33+3t22+3t
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B
2t33+3t22
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C
2t33+3t22+3t+2
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D
2t33+3t22−3t
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Solution
The correct option is C2t33+3t22+3t+2 Given, a(t)=4t+3 As we know that the velocity as a fuction of time is given by ∫vudv=∫t0(4t+3)dt ⇒[v]vu=[2t2+3t]t0 ⇒v−u=2t2+3t Or, ⇒v−3=2t2+3t⇒v(t)=2t2+3t+3 Now, final position of the particle is given as ∫xfxix(t)dt=∫t0v(t)dt=∫t0(2t2+3t+3)dt ⇒xf−xi=[2t33+3t22+3t]t0 ⇒xf−2=[2t33+3t22+3t] ⇒xf(t)=[2t33+3t22+3t+2]