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(B) Relative velocity of one train with respect to the other is $v_{rel} = 10 + 10 = 20 \, m/s$.Relative acceleration is $a_{rel} = 0.3 + 0.2 = 0.5 \,

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(B) Relative velocity of one train with respect to the other is $v_{rel} = 10 + 10 = 20 \, m/s$.Relative acceleration is $a_{rel} = 0.3 + 0.2 = 0.5 \, m/s^2$.To cross each other,the total distance to be covered is the sum of their lengths: $s = 100 + 125 = 225 \, m$.Using the equation of motion $s = ut + \frac{1}{2}at^2$:$225 = 20t + \frac{1}{2} 	imes 0.5 	imes t^2$$225 = 20t + 0.25t^2$$0.25t^2 + 20t - 225 = 0$Multiplying by $4$ to simplify: $t^2 + 80t - 900 = 0$.Solving the quadratic equation using the formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$t = \frac{-80 \pm \sqrt{6400 - 4(1)(-900)}}{2(1)}$$t = \frac{-80 \pm \sqrt{6400 + 3600}}{2} = \frac{-80 \pm \sqrt{10000}}{2} = \frac{-80 \pm 100}{2}$.Since time must be positive,$t = \frac{20}{2} = 10 \, s$.

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