A particle moves in a line with velocity given by dsdt=s+1. The time taken by the particle to cover a distance of 9 mts is

s=9 dcdt=s+1 ⇒∫dss+1=∫dt log s+1=t+c s=0 t=0 ⇒ c=0 s=9t=loge10

A particle moves in a line with velocity given by dsdt=s+1. The time taken by the particle to cover a distance of 9 mts is

s=9 dcdt=s+1 ⇒∫dss+1=∫dt log s+1=t+c s=0 t=0 ⇒ c=0 s=9t=loge10

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A particle moves in a line with velocity given by $\frac{ds}{dt} = s + 1$ . The time taken by the particle to cover a distance of 9 mts is

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$\log_{e} 10$

$2 \log_{e} 10$

$\log_{10} e$

answer is B.

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$s = 9 \frac{dc}{dt} = s + 1 \Rightarrow \int \frac{ds}{s + 1} = \int dt \log |s + 1| = t + c s = 0 t = 0 \Rightarrow c = 0 \begin{matrix} s = 9 \\ t = \log_{e} 10_{} \end{matrix}$

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A particle moves in a line with velocity given by dsdt=s+1. The time taken by the particle to cover a distance of 9 mts is

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A particle moves in a line with velocity given by $\frac{ds}{dt} = s + 1$ . The time taken by the particle to cover a distance of 9 mts is