A train moving with a speed of 60 km/hr is slowed down uniformly to 30 km/hr for repair purposes during running. After this it was accelerated uniformly to reach to its original speed. If the distance covered during constant retardation be 2 km and that covered during constant acceleration be 1 km, find the time lost in the above journey.

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Detailed Solution

Time taken by train in retardation
By using $s = (\frac{u + v}{2}) t \Rightarrow 2 = (\frac{60 + 30}{2}) t_{1} \Rightarrow t_{1} = \frac{2}{45} hr$
Time taken by train in accelerated motion
By using $(\frac{u + v}{2}) t \Rightarrow 1 = (\frac{30 + 60}{2}) t_{2} \Rightarrow t_{2} = \frac{1}{45} h$
Time taken by train to travel 3 km by uniform motion
$t_{3} = \frac{3}{60} = \frac{1}{20} hr$
Time lost = $(\frac{3}{45} - \frac{1}{20}) \times 60 = 1 \min$