A particle moves in a straight line with velocity given by $\frac{dy}{dx} = x+ 1$ (x being the distance described). The time taken by the particle to describe 99 metres is :

$2 \log _e 10$

$\frac{d x}{d t}=x+1 \Rightarrow \frac{d x}{x+1}=d t$

⇒ log(x + 1) = t + c             ....(1)

Initially, when t = 0, x = 0

∴ c = 0

∴ log (x + 1) = t

When x = 99, then t = log_e (100) = 2 log_e 10

Hence (2) is the correct answer.