Question

# The acceleration 'a' in $$m/s^2$$ of a particle is given by a $$a = 3t^2 + 2t + 2$$ where t is the time. If the particle starts out with a velocity u = $$2 m/s$$ at $$t=0$$, then the velocity at the end of 2 second is

A
12m/s
B
18m/s
C
27m/s
D
36m/s

Solution

## The correct option is B $$18 m/s$$We have, $$a=\dfrac{dV}{dt}$$Given $$a=3t^2+2t+2$$$$dV=a\times dt$$$$dV=(3t^2+2t+2)dt$$Integrating bothe sides, we get$$V=\dfrac{3\times t^3}{3}+\dfrac{2t^2}{2}+t=t^3+t^2+2t+C$$     .....(1)where $$C$$ is constantPutting $$t=0$$ in equation (1), we get$$2=0+C$$Then,$$C=2$$Substituting the value in equation (2), we get$$V=t^3+t^2+2t+2$$At $$t=2$$$$V=8+4+4+2=18\,m/s$$Physics

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