Question

A train is travelling at a speed of
90 km h –1 . Brakes are applied so
as to produce a uniform m
acceleration of – 0.5 m s -2 . Find
how far the train will go before it
is brought to rest.
ALSO EXPLAIN THE STEPS
GIVE CORRECT ANSWER

Answer 1

Answer:

625 metres

Explanation:

Given:

• Initial velocity of train = u = 90 km/h

90 km/h = 90×5/18 = 25 m/s

• Acceleration = a = - 0.5 m/s²
• Final velocity = v = 0 m/s

To find:

• Distance travelled by the train (s)

Using third equation of motion :

V²-u²=2as

0²-25²=2×-0.5×s

0-625= - 1s

-625 = - 1s

s = - 625/-1

s = 625/1

s=625 metres

The train will travel a distance of 625 metres before coming to rest

Answer 2

Answer:

Explanation:

Given:-

Initial speed of the train, u = 90 km/h = 25 m/s

The final speed of the train, v = 0

Acceleration = - 0.5 m/s²

To Find:-

Distance acquired

Formula to be used:-

Third equation of motion:

v² = u² + 2 as

Solution:-

Putting all the value, we get

v² = u² + 2as

⇒ (0)² = (25)² + 2 × (- 0.5) s

⇒ s = 625 m

Hence, The train will cover a distance of 625 m before coming to rest.