A particle is moving in a straight line. Its retardation is given as $$a=$$−$$loge2$$ V, here V is the speed of the particle in $$m/s$$. Find the time (in$$ $$s) in which speed of the particle reduces to half.

$$dVdt=$$−(loge2)V$$⇒ $$VV/2dVV=$$−$$(loge2) $$0tdt$$⇒[log V]$$VV/2=$$−log $$2$$ [t]⇒$$logV2$$−log $$V=$$−log $$21/2$$ [t]⇒$$log12=$$−(12log$$ $$2) t⇒−$$log2=$$−(12log$$ $$2) t⇒$$t=2s$$

Rectilinear Motion

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A particle is moving in a straight line. Its retardation is given as $a = - (\log_{e} \sqrt{2})$ V, here V is the speed of the particle in m/s. Find the time (in s) in which speed of the particle reduces to half.

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Solution

$\begin{array}{l} \begin{array}{l} \frac{d V}{d t} = - (\log_{e} \sqrt{2}) V \\ \Rightarrow_{V}^{V / 2} \frac{d V}{V} = - (\log_{e} \sqrt{2})_{0}^{t} d t \\ \Rightarrow {[\log V]}_{V}^{V / 2} = - \log \sqrt{2} [t] \\ \Rightarrow \log \frac{V}{2} - \log V = - \log 2^{1 / 2} [t] \\ \Rightarrow \log \frac{1}{2} = - (\frac{1}{2} \log 2) t \\ \Rightarrow - \log 2 = - (\frac{1}{2} \log 2) t \end{array} \\ \Rightarrow t = 2 s \end{array}$

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Rectilinear Motion

Remember concepts with our Masterclasses.

A particle is moving in a straight line. Its retardation is given as a=−loge2 V, here V is the speed of the particle in m/s. Find the time (in s) in which speed of the particle reduces to half.

dVdt=−(loge2)V⇒ VV/2dVV=−(loge2) 0tdt⇒[log V]VV/2=−log 2 [t]⇒logV2−log V=−log 21/2 [t]⇒log12=−(12log 2) t⇒−log2=−(12log 2) t​⇒t=2s

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A particle is moving in a straight line. Its retardation is given as $a = - (\log_{e} \sqrt{2})$ V, here V is the speed of the particle in m/s. Find the time (in s) in which speed of the particle reduces to half.