Question

a train starts from rest and accelerates uniformly for 30sec to acquire a velocity of 108km/h . it travels with this velocity for 20 min . the driver applies the brake and the train returns uniformly to stop after 20 sec . Find the total distance covered by the train .

Answer 1

total time =20 min +20sec+20sec
time =1200 +20+20
time=1240sec
108km/h=108×5/18=30m/sec
d=s×t
d=1240×30
d=37200metre=37.2km
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Answer 2

Answer:

$36750 \: m$

Explanation:

When the train starts from rest,

u= 0,

t= 30s,

v= 108 km/h = 108 × 5/18 = 30 m/s.

Now,

a= v-u/t = 1 m/s².

Distance ,S1 = ut + 1/2 at²

= 0 + 1/2 × 1 × 30 × 30

= 450 m.

At uniform velocity of 30 m/s for 20 min, i.e., 1200s,

Distance, S2 = 30 × 1200 = 36000 m.

Now,

On applying brakes,

t= 20s,

u= 30 m/s,

v= 0.

a= v-u/t = 0-30/20 = -1.5 m/s²

Distance S3 = v² - u²/2a

= 0 - (30)² /2 × 1.5

= 300 m

Therefore,

The total distance traversed = S1 + S2 + S3

= (450 + 36000 + 300) m

= 36750 m