Question

A bullet train in Japan Starting from rest accelerates uniformly at 0.5 m/s² for 2 minutes. It then maintains uniform speed for 20 minutes and then retards uniformly at the rate of 0.25 m/s² and comes to rest. Calculate the average speed of the train.

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Answer 1

Answer:

$Average . . Speed$ $= 53.08 m/s$

Step-by-step explanation:

$Average . . Speed = \frac{Total Distance Covered}{Total Time Taken}$

From rest, accelerates uniformly at $0.5 m/s^2$ for 2 minutes = 120 seconds

Distance covered during this period, $s = ut + \frac{1}{2} at^2$  $= (0. 120 + \frac{1}{2} . 0.5 . 120^2) m$

= 3600 m

Final speed reached, $v = u + at$ = 0 + 0.5 x 120 = 60 m/s

Then maintains uniform speed for 20 minutes = 1200 seconds

Distance covered during this period, $s = v t$ = 60 x 1200 m= 72000 m

Then retards uniformly at the rate of $0.25 m/s^2$ and comes to rest

$v = u - at$ => 0 = 60 - 0.25t $=> t = \frac{60}{0.25} = 240 s$

Distance covered during this period, $s = ut - \frac{1}{2} at^2$  $= (60. 240 - \frac{1}{2} . 0.25 . 240^2) m$ =30 x 240 m = 7200 m

$Average . . Speed = \frac{Total Distance Covered}{Total Time Taken}$ $= \frac{3600 + 72000 + 7200}{120 + 1200 + 240}$ $= \frac{82800}{1560}$ $= 53.08 m/s$