If the velocity of a particle is $v=2A t+3B t^{2}$, where $A$ and $B$ are constants, then the distan

Question

If the velocity of a particle is $v=2A t+3B t^{2}$, where $A$ and $B$ are constants, then the distance travelled by it between $0\mathrm{~s}$ and $2\mathrm{~s}$ is

Professor PC Thomas Classes · Answer

(a) : Velocity of the particle is $v=2A t+3B t^{2}$ \ $\frac{d s}{d t}=2A t+3B t^{2} \implies \int d s=\int_{0}^{2}\left(2A t+3B t^{2}ight) d t$ \ $	herefore \quad s=\left[A t^{2} + B t^{3}ight]_{0}^{2}$ \ Required distance $= A(2)^{2} + B(2)^{3} - 0 = 4 A+8 B$