Question

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

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

Answer 1

Answer: (B)

$V = At + Bt^{2} \Rightarrow \frac{dx}{dt} =At + Bt^{2} $
$ \Rightarrow \int^{x}_{0} dx = \int^{2}_{1} \left(At + Bt^{2}\right)dt $
$ \Rightarrow x = \frac{A}{2}\left(2^{2} - 1^{2}\right) + \frac{B}{3}\left(2^{3} - 1^{3}\right) = \frac{3A}{2} + \frac{7B}{3} $