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If the velocity of a particle is $v = A t + Bt^{2}$, where A and B are constants, then the distance travelled by it between 1 s and 2 s is:
Explanation
Given: Velocity $V = A t + Bt^{2} \Rightarrow dx/dt = A t + Bt^{2}$ By integrating we get distance travelled $\Rightarrow \int_{0}^{x}{dx = \int_{1}^{2}{(At + Bt^{2})}dt}$ Distance travelled by the particle between 1 s and 2 s $x = A/2(2^{2} - 1^{2}) + B/3(2^{3} - 1^{3}) = 3A/2 + 7B/3$