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

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$\frac{A}{2} + \frac{B}{3}$

$\frac{3}{2} A + 4 B$

$\frac{3}{2} A + \frac{7}{3} B$

$3 A + 7 B$

answer is C.

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Detailed Solution

$\int_{0}^{x} d x = \int_{1}^{2} V (d t)$

$\int_{0}^{x} d x = \int_{1}^{2} (A t + B t^{2}) (d t)$

$x = \frac{A}{2} (2^{2} - 1^{2}) + \frac{B}{3} (2^{3} - 1^{1}) = \frac{3}{2} A + \frac{7}{3} B$

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