Acceleration of the particle in a rectilinear motion is given as a=2t−6 in (m/s2). The displacement as a function of time t respectively is
(Given, at t=0,u=2m/s,x=2m)
A
2t33−3t2+2t+2
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B
t33−3t2+2t
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C
t33−3t2+2t+2
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D
t33−3t2
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Solution
The correct option is Bt33−3t2+2t Given, a(t)=2t−6
As we know that the velocity as a fuction of time is given by ∫vudv=∫t0(2t−6)dt ⇒[v]vu=[t2−6t]t0 ⇒v−u=t2−6t ⇒v−2=t2−6t⇒v(t)=t2−6t+2
Now, displacement is given by x(t)=∫t0v(t)dt=∫t0(t2−6t+2)dt ⇒x(t)=[t33−3t2+2t]t0 ⇒x(t)=(t33−3t2+2t)