A train is moving on straight track with velocity v₀ = 13.5 ms^-1. To stop the train at a particular station, the driver applies brakes at t = 0, which is caused of a retardation proportional to the velocity of the train. The speed of train reduces 50% in the first 2 s in t₀=4 . The velocity of train (in ms^-1) at t=4s (Given , e=2.7)is

Difficult

Solution

$∵ a = - k v or \frac{d v}{d t} = - k v$
$\begin{array}{l} Or \int_{v_{0}}^{v} \frac{d v}{v} = - k \int_{0}^{t} d t or [\ln v]_{v_{0}}^{v} = - k t \\ Or In v - \ln v_{0} = - k t or \ln \frac{v}{v_{0}} = - k t \\ Or \frac{v}{v_{0}} = e^{- k t} \end{array}$
$\begin{array}{l} For t = t_{0}, v = v_{0} - v_{0} \times \frac{50}{100} = \frac{v_{0}}{2} \\ ∴ \frac{v_{0}}{2} = v_{0} e^{- k t} \end{array}$
$\begin{array}{l} Or v = v_{0} e^{\frac{- \ln 2}{t_{0}} t} or \ln v = \ln v_{0} - \frac{\ln 2}{t_{0}} t \\ Or \ln \frac{v}{v_{0}} = - \frac{\ln 2}{4 \ln 2} t or \ln \frac{v}{v_{0}} = - \frac{t}{4} or \frac{v}{v_{0}} = e^{- \frac{t}{4}} \\ ∴ v = v_{0} e^{- \frac{t}{4}} = 13.5 e^{- \frac{4}{4}} = \frac{13.5}{e} = \frac{13.5}{2.7} = 5 {ms}^{- 1} \end{array}$

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