Q 11-02-112JEE MainJEE Main 2021 (25 Jul, Shift 2)Easy
The instantaneous velocity of a particle moving in a straight line is given as $v = \alpha t + \beta t^2$, where $\alpha$ and $\beta$ are constants. The distance travelled by the particle between $1$ s and $2$ s is:
Answer: (B) $\frac{3}{2}\alpha + \frac{7}{3}\beta$
The velocity does not change sign in this interval, so
$$s = \int_1^2(\alpha t + \beta t^2)\,dt = \frac{\alpha}{2}(4 - 1) + \frac{\beta}{3}(8 - 1) = \frac{3}{2}\alpha + \frac{7}{3}\beta$$
Solution by Sreeraj P, M.Sc Physics