Acceleration of the particle in a rectilinear motion is given as a=2t−2 in (m/s2). The final position of the particle as a function of time t is
(Given that at t=0,u=2m/s,x=4m)
A
t33−t2+2t+4
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B
t33−t2+2t
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C
t33−t2+4
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D
t33−t2
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Solution
The correct option is At33−t2+2t+4 Given, a(t)=2t−2
As we know that the velocity as a fuction of time is given by ∫vudv=∫t0(2t−2)dt ⇒[v]vu=[t2−2t]t0 ⇒v−u=t2−2t
Or, ⇒v−2=t2−2t⇒v(t)=t2−2t+2
Now, final position of the particle is given as ∫xfxix(t)dt=∫t0v(t)dt=∫t0(t2−2t+2)dt ⇒xf−xi=[t33−t2+2t]t0 ⇒xf−4=[t33−t2+2t] ⇒xf(t)=[t33−t2+2t+4]