If the velocity of a particle is v = At + Bt², where A and B constants, then the distance travelled by it between 1s and 2s is.

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

$3 A + 7 B$

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

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

answer is C.

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Detailed Solution

V = At + Bt²
$\frac{ds}{dt} = At + {Bt}^{2}$

Integrating both sides we get
$\begin{array}{l} \int ds = \int_{1}^{2} (At + {Bt}^{2}) dt \\ S = {[\frac{{At}^{2}}{2} + \frac{{Bt}^{3}}{3}]}_{1}^{2} = \frac{4 A}{2} + \frac{8 B}{3} - \frac{A}{2} - \frac{B}{3} \\ = 3 \frac{A}{2} + \frac{7 B}{3} \end{array}$