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

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

3A +7B

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

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

answer is C.

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Detailed Solution

V = At+ ${Bt}^{2} \Rightarrow \frac{dx}{dt} = At + {Bt}^{2}$

$\Rightarrow \int_{0}^{x} dx = \int_{1}^{2} (At + {Bt}^{2}) dt$

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

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