Acceleration of the particle in a rectilinear motion is given as a=4t+3 in (m/s2). The velocity and displacement as a function of time t respectively are
(Given that at t=0,u=3m/s,x=0m)
A
(t2+3),(t33+3t22+3t)
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B
(t2+3t),(t33+3t22)
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C
(2t2+3t+3),(2t33+3t22+3t)
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D
(2t2+3t),(2t33+3t22)
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Solution
The correct option is C(2t2+3t+3),(2t33+3t22+3t) 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, displacement is given by x(t)=∫t0v(t)dt=∫t0(2t2+3t+3)dt ⇒x(t)=[2t33+3t22+3t]t0 ⇒x(t)=[2t33+3t22+3t]