A train starting from rest is accelerated and the instantaneous acceleration is given by 10v+1m/s2, where ‘v’ is the velocity in m/s. Find the distance (in m) in which the train attains a velocity of 54 km/hr.

Given v=54km/hr=54×518=15m/s a=10v+1⇒dvdt=10v+1 dvdt×dxdx=10v+1⇒vdvdx=10v+1 ∫015v(v+1)dv=∫0x10dx (v33+v22)015=10x⇒x=123.75m

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A train starting from rest is accelerated and the instantaneous acceleration is given by $\frac{10}{v + 1} m / s^{2},$ where ‘v’ is the velocity in m/s. Find the distance (in m) in which the train attains a velocity of 54 km/hr.

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answer is 123.75.

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Detailed Solution

Given $v = 54 k m / h r = 54 \times \frac{5}{18} = 15 m / s$
$a = \frac{10}{v + 1} \Rightarrow \frac{d v}{d t} = \frac{10}{v + 1}$
$\frac{d v}{d t} \times \frac{d x}{d x} = \frac{10}{v + 1} \Rightarrow v \frac{d v}{d x} = \frac{10}{v + 1}$
$\int_{0}^{15} v (v + 1) d v = \int_{0}^{x} 10 d x$
${(\frac{v^{3}}{3} + \frac{v^{2}}{2})}_{0}^{15} = 10 x \Rightarrow x = 123.75 m$

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