### NEET Solved Paper 2016 Question 41

##### 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 :-

#### Options:

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} $

## Show Answer

#### Answer:

Correct Answer: C

#### Solution:

- $ V=At+Bt^{2}\Rightarrow \frac{dx}{dt}=At+Bt^{2} $

$ \Rightarrow $ $ \int\limits_0^{x}{dx}=\int\limits_2^{2}{(At+Bt^{2})dt} $

$ \Rightarrow $ $ x=\frac{A}{2}(2^{2}-1^{2})+\frac{B}{3}(2^{3}-1^{3})=\frac{3A}{2}+\frac{7B}{3} $