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) $v_0 + \dfrac{g}{2} + \dfrac{f}{3}$
Concept. Displacement is the integral of velocity, $x=\displaystyle\int_0^t v\,dt$, with $x=0$ at $t=0$.
Step 1. Integrate.
$$x=\int_0^1(v_0+gt+ft^2)\,dt=\left[v_0t+\frac{gt^2}{2}+\frac{ft^3}{3}\right]_0^1$$
Step 2. Evaluate. $x=v_0+\dfrac g2+\dfrac f3$.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
