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(B) Given the differential equation: $\frac{dx}{dt} = x + 1$. Separating the variables,we get: $\frac{dx}{x + 1} = dt$. Integrating both sides: $\int

Vedclass · Accepted Answer

$2 \log_e 10$

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(B) Given the differential equation: $\frac{dx}{dt} = x + 1$. Separating the variables,we get: $\frac{dx}{x + 1} = dt$. Integrating both sides: $\int \frac{dx}{x + 1} = \int dt$. This yields: $\ln(x + 1) = t + C$. At $t = 0$,the distance $x = 0$. Substituting these values: $\ln(0 + 1) = 0 + C \implies \ln(1) = C \implies C = 0$. Thus,the equation for time is: $t = \ln(x + 1)$. To find the time taken to cover a distance of $x = 99 \ m$: $t = \ln(99 + 1) = \ln(100)$. Since $\ln(100) = \log_e(10^2) = 2 \log_e 10$. Therefore,the correct option is $B$.

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

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