Question

If the velocity of a particle is $v=At+B{t}^2$, where $A$ and $B$ are constants, then the distance travelled by it between $1s$ and $2s$ is

• A. $\frac{3}{2}A+4B$
• B. $3A+7B$
• C. $\frac{3}{2}A+\frac{7}{3}B$
• D. $\frac{A}{2}+\frac{B}{3}$

Answer 1

Answer: (C)

$\frac{3}{2}A+\frac{7}{3}B$

We know that $v=\frac{dS}{dt}$

Integrating, $S={\int }_{t=1}^{t=2}vdt={\int }_1^2(At+B{t}^2)dt=A/2(2^2-1)+B/3(2^3-1)=\frac{3}{2}A+\frac{7}{3}B$