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NEET Physics QUESTION #7321
Question 1

The velocity of a particle is given by $v = At + Bt^2$, where $A$ and $B$ are constants. The distance travelled between $t = 1\text{ s}$ and $t = 2\text{ s}$ is:

  • $\dfrac{3}{2}A + 4B$
  • $3A + 7B$
  • $\dfrac{3}{2}A + \dfrac{7}{3}B$✔️
  • $\dfrac{A}{2} + \dfrac{B}{3}$
Correct Answer Explanation

$x = \displaystyle\int_1^2 v\,dt = \int_1^2 (At + Bt^2)\,dt = \left[\dfrac{At^2}{2} + \dfrac{Bt^3}{3}\right]_1^2$

$= \left(\dfrac{4A}{2} + \dfrac{8B}{3}\right) - \left(\dfrac{A}{2} + \dfrac{B}{3}\right) = \mathbf{\dfrac{3A}{2} + \dfrac{7B}{3}}$