A particle moves in a straight line with a velocity given by dxdt=x+1 (x is the distance travelled). The time taken by a particle to traverse a distance of 99 metres is

dxdy=x+1⇒ ∫1x+1 dx= ∫dt⇒ log (x+1)=t+cPut t=0, x=0⇒ c= log(1)=0∴ t= log(x+1)Put x=99t= log 100 = log 102=2 log 10

A particle moves in a straight line with a velocity given by dxdt=x+1 (x is the distance travelled). The time taken by a particle to traverse a distance of 99 metres is

dxdy=x+1⇒ ∫1x+1 dx= ∫dt⇒ log (x+1)=t+cPut t=0, x=0⇒ c= log(1)=0∴ t= log(x+1)Put x=99t= log 100 = log 102=2 log 10

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

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$1 / 2 \log_{10} e$

$2 \log_{e} 10$

$\log_{10} e$

$2 \log_{10} e$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{d x}{d y} = x + 1 \\ \Rightarrow \int \frac{1}{x + 1} d x = \int d t \\ \Rightarrow \log (x + 1) = t + c \\ P u t t = 0, x = 0 \\ \Rightarrow c = \log (1) = 0 \\ ∴ t = \log (x + 1) \\ P u t x = 99 \\ t = \log 100 = \log 10^{2} = 2 \log 10 \end{array}$