A train is travelling at a speed of 60 km/ h. Brakes are applied so as to produce a uniform acceleration of −0.5 m /s2. Find how far the train will go before it is brought to rest.
Answers
Question:
A train is travelling at a speed of 60 km/ h. Brakes are applied so as to produce a uniform acceleration of −0.5 m /s2. Find how far the train will go before it is brought to rest.
Formula Used:
- v = u + at
- S = ut + 1/2at²
Solution:
Speed of train , u = 90 km/h = 90 × 5/18 = 25 m/s
acceleration , a = -0.5 m/s²
Finally train will be rest so, final velocity, v = 0
0 = 25 - 0.5t
25 = 0.5t ⇒t = 50 sec
- S = at + 1/2at²
Where S is distance travelled before stop
S = 25 × 50 - 1/2 × 0.5 × 50²
= 1250 - 1/2 × 0.5 × 2500
= 1250 - 625
= 625 m
Hence, distance travelled = 625m and time taken = 50 sec
Answer:
hope this helps you
Explanation:
et the initial speed of the train be u=90Km/h
=25m/s.
Final speed of the train, v=0 (train comes to rest)
Acceleration a=0.5ms
−2
As per 3
rd
law of motion,
v
2
=u
2
+2as
(0)
2
=(25)
2
+2(0.5)s
s= train travelled distance.
∴s=
2(0.5)
(25)
2
=625m.
Train will travel 625km before it is brought to rest.