Question

A train is travelling at a speed of 90 km/h brakes are applied so as to produce a uniform acceleration of -0.5 m s-2 find how far the train will go before it is brought to rest?

Answer 1

Speed of train , u = 90 km/h = 90 × 5/18 = 25 m/s
acceleration , a = -0.5 m/s²
Use formula,
v = u + at
Finally train will be rest so, final velocity, v = 0
0 = 25 - 0.5t
25 = 0.5t ⇒t = 50 sec

Again, use formula,
S = ut + 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 2

Answer:

Explanation:

Given :-

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

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

⇒ v = u2 + 2 as

⇒ (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.