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(B) The velocity of the particle is given by $v = At + Bt^2$.Since $v = \frac{ds}{dt}$,we have $ds = (At + Bt^2) dt$.To find the distance travelled be

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

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(B) The velocity of the particle is given by $v = At + Bt^2$.Since $v = \frac{ds}{dt}$,we have $ds = (At + Bt^2) dt$.To find the distance travelled between $t = 1 \ s$ and $t = 2 \ s$,we integrate the velocity with respect to time:$s = \int_{1}^{2} (At + Bt^2) dt$$s = \left[ \frac{At^2}{2} + \frac{Bt^3}{3} \right]_{1}^{2}$$s = \left( \frac{A(2)^2}{2} + \frac{B(2)^3}{3} \right) - \left( \frac{A(1)^2}{2} + \frac{B(1)^3}{3} \right)$$s = \left( 2A + \frac{8B}{3} \right) - \left( \frac{A}{2} + \frac{B}{3} \right)$$s = (2A - \frac{A}{2}) + (\frac{8B}{3} - \frac{B}{3})$$s = \frac{3}{2}A + \frac{7}{3}B$Since the velocity does not change sign in the interval $[1, 2]$,the distance is equal to the magnitude of displacement.

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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