Q 11-02-019JEE MainAIEEE 2007Medium
The velocity of a particle is $v = v_0 + gt + ft^2$. If its position is $x = 0$ at $t = 0$, then its displacement after unit time ($t = 1$) is
Answer: (B) $v_0 + \dfrac{g}{2} + \dfrac{f}{3}$
Displacement is the integral of velocity:
$$x = \int_0^1 (v_0 + gt + ft^2)\,dt = \left[v_0t + \frac{gt^2}{2} + \frac{ft^3}{3}\right]_0^1 = v_0 + \frac{g}{2} + \frac{f}{3}$$
Solution by Sreeraj P, M.Sc Physics