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Question

If the velocity of a particle is $$v=At+Bt^2$$, where $$A$$ and $$B$$ are constants, then the distance travelled by it between $$1\ s$$ and $$2\ s$$ is


A
32A+4B
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B
3A+7B
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C
32A+73B
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D
A2+B3
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Solution

The correct option is B $$\displaystyle\dfrac{3}{2}A+\displaystyle\dfrac{7}{3}B$$
We know that $$v=\dfrac{dS}{dt}$$

Integrating, $$S=\int_{t=1}^{t=2}vdt=\int_1^2 (At+Bt^2)dt=A/2(2^2-1)+B/3 (2^3-1)=\dfrac{3}{2}A+\dfrac{7}{3}B$$

Physics

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