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A metro train starts from rest and in 5 s achieves $108 km/h$. After that it moves with constant velocity and comes to rest after travelling 45 m with uniform retardation. If total distance travelled is 395 m, find total time of travelling.
Explanation
Given: $u = 0$, $t = 5\text{sec}$, $v = 108 km/hr = 30 m/s$ By $eq ^{n}$ of motion $v = u + a t$ or $a = v/t = 30/5 = 6 m/s^{2} [ \because u = 0]$ $S_{1} = \frac{1}{2}a t^{2} = \frac{1}{2} \times 6 \times 5^{2} = 75 m$ Distance travelled in first 5sec is 75 m. Distance travelled with uniform speed of $30 m/s$ is $S_{2}$ $395 = S_{1} + S_{2} + S_{3} \Rightarrow 395 = 75 + S_{2} + 45 \Rightarrow S_{2} = 275 m$ Time taken to travel $275 m = 275/30 = 9.2\text{sec}$ For retarding motion, we have $0^{2} - 30^{2} = 2(-a) \times 45$, we get $a = 10 m/s^{2}$ $S = ut + \frac{1}{2}a t^{2} \Rightarrow 45 = 30t + \frac{1}{2}(-10)t^{2} \Rightarrow 45 = 30t - 5t^{2}$ On solving we get, $t = 3\text{sec}$ Total time taken = $5 + 9.2 + 3 = 17.2\text{sec}$.