A train starts from rest and accelerates uniformly for 30 seconds to acquire a Velocity of 108 km/h. It travels with this velocity for 20 minutes. The driver now applies brakes and the train slows down uniformly to stop after 20 seconds. Find the total distance covered by the train.
Answers
♨ Answer:-
36750 m.
♨ Step-by-step 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. [The required solution..]
Answer:
36750 m.
♨ Step-by-step 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. [The required solution..]
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