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 s2. Find
how far the train will go before it
is brought to rest.​

Answer 1

Answer:initial speed, u = 90 km/h = 90 x (1000/3600) m/s = 25 m/s

final speed, v  = 0 m/s

 

uniform acceleration, a = -0.5 m/s2

 

Using equation of motion: v2 = u2 + 2as

We have:                         0  =  (25)^2  +  2x(-0.5)xs

                                      s = 625 m                          

[s: distance traveled by the train after the brakes are applied and before it is finally brought into rest]

Explanation:

Answer 2

Answer:

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  the distance travelled before stopping

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

                          OR

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

⇒ 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.

BOTH ARE CORRECT ANSWER