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The maximum acceleration or deceleration that a train may have is a = 5 ms^-2. The minimum time in which the train may reach from one station to the other separated by a distance d=500 m is t₀=5 n s. The value of n is _______.

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answer is 4.

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$\begin{array}{l} d = \frac{1}{2} \times v_{0} t_{0} = \frac{v_{0} t_{0}}{2} \\ and \tan θ = a = \frac{v_{0}}{t_{0} / 2} = \frac{2 v_{0}}{t_{0}} \end{array}$

$\begin{array}{l} ∴ v_{0} = \frac{a t_{0}}{2} \\ ∴ d = \frac{t_{0}}{2} \times \frac{a t_{0}}{2} = \frac{a t_{0}^{2}}{4} \\ ∴ t_{0} = 2 \sqrt{\frac{d}{a}} = 2 \sqrt{\frac{500}{5}} = 20 s = 20 = 5 n \\ ∴ n = 4 \end{array}$

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