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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
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B
18m/s
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C
27m/s
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D
36m/s
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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 constant

Putting $$t=0$$ 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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