4. A train starts from rest and accelerates uniformly for 30s to acquire a
velocity of 108km/h. It travels with this velocity for 20min. The driver
now applies brakes and the train retards uniformly to stop after 20s.
Find the total distance covered be the train.
Answers
Answer:
ANSWER:
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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Answer:
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