Q. The velocity of 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
A) $v_{0} + 2g + 3f$
B) $v_{0} + g/2 + f/3$
C) $v_{0} + g + f$
D) $v_{0} + g/2 + f$
Explanation
(b) $v = v_{0} + gt + ft^{2}$
Or $\frac{dx}{dt} = v_{0} + gt + ft^{2}$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dx = \left( v_{0} + gt + ft^{2} \right)\ dt$
So, $\ \ \ \ \ \ \ \ \ \ \int_{0}^{x}{dx} = \int_{0}^{1}{\left( v_{0} + gt + ft^{2} \right)\ dt}$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = v_{0} + \frac{g}{2} + \frac{f}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.