A particle moves in a straight line with a velocity given by dx-dt - x - 1 x is the distance described- The time taken by a particle to traverse a distance of 99 metres is-

The correct option is B 2 log_e 10 dx/dt=x+1 #160;∫_0^99dx/x+1=∫_0^t dt ln(x+1) ∫_0^99=t ln 100 #160;=t t=2 #160;ln #160 ...

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Standard XII

Mathematics

Average Rate of Change

A particle mo...

Question

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

l o g_{1} e

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2 l o g_{e} 10

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2 l o g_{10} e

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\frac{1}{2} l o g_{10}

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Solution

The correct option is B $2 l o g_{e} 10$
$\frac{d x}{d t} = x + 1$
$\int_{0}^{99} \frac{d x}{x + 1} = \int_{0}^{t} d t$
$l n (x + 1) \int_{0}^{99} = t$
$l n 100 = t$
$t = 2 l n 10$

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

The correct option is B 2loge10dxdt=x+1∫990dxx+1=∫t0dtln(x+1)∫990=tln100=tt=2ln10

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

The correct option is B $2 l o g_{e} 10$
$\frac{d x}{d t} = x + 1$
$\int_{0}^{99} \frac{d x}{x + 1} = \int_{0}^{t} d t$
$l n (x + 1) \int_{0}^{99} = t$
$l n 100 = t$
$t = 2 l n 10$