Question

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

• A. $\log_{10}e$
• B. $2\log_{e}10$
• C. $\log_{10}e$
• D. $\frac{1}{2}\log_{10}e$

Answer 1

Answer: (B)

$2\log_{e}10$

Answer 2

We have $\frac{dx}{x + 1} = dt$

Integrating, $\int_{0}^{99}\frac{dx}{x + 1} = \int_{0}^{t}{dt}$ ⇒ $\lbrack\ln(x + 1)\rbrack_{0}^{99} = t$

∴ $t = \ln 100 = \log_{e}(10)^{2} = 2\log_{e}10$