Question

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:

• A. $v_0+2g+3f$
• B. $v_0+\frac{g}{2}+\frac{f}{3}$
• C. $v_0+g+f$
• D. $v_0+\frac{g}{2}+f$

Answer 1

The correct option is B $v_0+\frac{g}{2}+\frac{f}{3}$

Solution 2

We know that $\frac{dx}{dt}=v$ or $dx=vdt$

so on integrating it we get ${\int }_{x=0}^{X}dx={\int }_{t=0}^{t=1}vdt$

or $X-0=v_0(1-0)+\frac{g}{2}(1^2-0)+\frac{f}{3}(1^3-0)$

or displacement for interval between $t=0$ and $t=1s$ will be $X=v_0+\frac{g}{2}+\frac{f}{3}$